discrete-math

[spring hound]

OK, then you need the not equal part.

[spring quartz]

so: a childOf p ^ b childOf q a | a not equal to b?

[spring hound]

What is the "a |" for?

[spring quartz]

oh i just wrote that at the end so it wouldnt get confusing in the sentence for me

[spring hound]

Well, you should and them all together.

[\exists p, q \in \text{Person}. p \text{ siblingOf } q \land a \text{ childOf } p \land b \text{ childOf } q \land a \ne b]

[vital dew]

Chai T. Rex

[spring hound]

Does that make sense?

[spring quartz]

yess

[spring hound]

OK, now we only need to say that (a \text{ cousinOf } b) is equivalent to that.

[vital dew]

Chai T. Rex

[spring hound]

How do we do that?

[spring quartz]

a cousinOf b <=> there exists p, q which are elements of Person, p siblingOf q ^ a childOf p ^ b childOf q ^ a and b are not the same person

[spring hound]

OK, but make sure to use the double lined arrow, like <=>.

I think that <-> is kind of like asking whether they're equivalent. <=> is stating that they're equivalent.

[spring quartz]

ohh i see

thank youu

thank u for being so patient with me 😭

im terrible at maths but good at at the more programming side of computer science

[spring hound]

You're welcome.

[spring quartz]

it mustve been frustrating sometimes and im sorry for that

[spring hound]

No problem.

[fossil mural]

to respond to this:
technically that is what you would do, just, having multiple of the same quantifier right next to each other turns out to be so common that it's useful to abbreviate it

so $$\exists p,q \in \text{Person}.$$ is just a shorter way of writing $$\exists p \in \text{Person}.\exists q \in \text{Person}.$$

[vital dew]

bee [it/its]

[spring quartz]

ohhh

thank youu

[acoustic pond]

hey I need to write the negation of this term:

If M is a set and R ⊆ M × M is a relation on M, then
R is called dichotomous if for all a, b ∈ M with a != b it holds:
aRb ⇐⇒ ¬(bRa).

Define the relation property non-dichotomously as
Negation of dichotomous. Just like for dichotomous only Elements with a != b are considered. Form the expression
as far as possible so that any negative signs that occur are as wide as possible
to the right of your printout.

How can I write the non-negated version and then the negated verson?

non-negated: ∀a∈M,∀b∈M(a≠b⇒(aRb⇔¬(bRa)))

negated: ∃a∈M,∃b∈M(a≠b∧((aRb∧bRa)∨(¬(aRb)∧¬(bRa))))

is this correct?

[stray reef]

a 6= b
what

@acoustic pond did you try to OCR this or copy-paste it from a PDF or something

[little prism]

iirc its their version of != for some reason

[acoustic pond]

edited it sry

[spring hound]

Checked it again and:
¬∀a∈M,∀b∈M(a≠b⇒(aRb⇔¬(bRa)))
∃a∈M,¬∀b∈M(a≠b⇒(aRb⇔¬(bRa)))
∃a∈M,∃b∈M¬(a≠b⇒(aRb⇔¬(bRa)))
∃a∈M,∃b∈M¬(¬(a≠b)∨(aRb⇔¬(bRa)))
∃a∈M,∃b∈M(a≠b∧¬(aRb⇔¬(bRa)))
∃a∈M,∃b∈M(a≠b∧(aRb⇔bRa))
∃a∈M,∃b∈M(a≠b∧((aRb∧bRa)∨(¬(aRb)∧¬(bRa))))

I like the next to last line as a simpler form, though.

[acoustic pond]

that looks great 😄 thank you!

[spring hound]

No problem.

[acoustic pond]

For all elements of the domain there is an element of the range of values ​​and for all elements of the range of values ​​there is an element of the domain.

D = domain
W = range of values

is this correct?: "∀x∈D,∃y∈W:∃z∈D:(y=f(x)∧x=f^−1 (y))"

[spring hound]

Well, f^-1 might not exist. For example, with f(x) = 5, there is no inverse of f.

You need to state it like y = f(x) or something.

But, there's a more fundamental problem.

It says for all elements of the range, but you don't have a forall v in W statement.

I'd recommend splitting it into two statements: one statement for the domain -> range and one statement for the range -> domain. Then, and those together.

So, you shouldn't have a bunch of quantifiers and then try to write the statement from inside those quantifiers.

Instead, you should have forall x in D (...) AND forall x in W (...).

If that's too much, you can do the course challenge on Khan Academy's Algebra 1 course (https://www.khanacademy.org/math/algebra), then work on the parts you had trouble with inside Khan Academy.

[acoustic pond]

ty I will look that up later

[void zinc]

Guys is this valid

No X are Y

all Z are X

some Z are not Y

it should be all Z are not Y

But would that still hold

[rich saffron]

You could run into an issue if Z is an empty set

But if there's at least one Z then that would hold

[mental pecan]

Good luck

Study hard

[cerulean radish]

How many triplets $(x_1, x_2, x_3) \in { 1, \dots, 55 }^3$ are there with $x_1 + x_2 + x_3 = 111$?

[vital dew]

A Lonely Bean

[haughty garden]

this smells like ||stars and bars||

[cerulean radish]

How will that look when the integers have an upper bound?

I haven't really dealt with this type of questions

[stray reef]

unga bunga is using incl excl on this

[cerulean radish]

I think I got it, thanks

[vital dew]

Kevin

[solar marsh]

.

[cerulean radish]

What do the double parenthesis mean?

[cerulean radish]

I see

[solar marsh]

combinatorics with repetition

[cerulean radish]

We haven't really covered that in the course but thank you anyway, I might want to look into combinatorics with repetition before I encounter another problem of this type

[fresh hull]

Execution time will be crazy

[errant bear]

crazy fast? its iterating through only 175k with a single boolean check

yeah its like 0.01 secs on a potato online compiler

[cerulean radish]

The execution time is constant

There are no inputs involved

[meager prawn]

Unfortunately, asymptotic behavior only matters so much in reality

[solar marsh]

2seg

[fresh hull]

I mean this is how I'd do it lol

but can there be something more efficient

[scarlet heath]

how would you read this out loud?

[brave rock]

"The power set of the union of the A_i is not a subset of the union of the power sets of A_i"

[scarlet heath]

you'd just call that bit A_i?

[brave rock]

the A_i s

however you want to call them

[scarlet heath]

I wasn't sure if you might have to say something about indexed families and index sets for the A_i

[brave rock]

I really don't think its worth the effort to say how it's indexed

Usually there's only one index set.

[scarlet heath]

got it. I am learning from a book instead of a class so I am not sure what's standard when people are actually discussing these topics

thank you!

[tardy violet]

puiiai is not a subset of uiipai 😎

[scarlet heath]

math school day 1 I will say this

[dire stag]

Why is 13 b. "true or false" instead of just "false"?

same for c, why is it "false" when it could be "true or false"?

12. Makes perfect sense, but 13, not so much

[errant bear]

if 12 makes sense then 13 should make sense

[errant bear]

why do you think 13b should be false and 13c should be true or false?

[dire stag]

Oh wait...

Yeah I see it now. My bad, first day studying Discrete.

I should probably invest more time into the rubber ducky method.

Just me typing out my thought process was enough to give me an answer.

thanks!

[snow sleet]

@solar marsh already answered your question but I'll supply you with the way I counted this up using a generating function. It's prolly beyond your discrete course (was definitely beyond mine) but it might peak your interest. Tagging you, Kevin, here as well to show how to count this without using your code. Although, your code is certainly another great way to check answers!

,w Expand[(Sum[x^i,{i,1,55}])^3]

[vital dew]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by logician1

[snow sleet]

look at the coefficient of x^(111). That coefficient is your answer.

Notice that I've expanded (x^1+x^2+x^3+...+x^55)(x^1+x^2+x^3+...+x^55)(x^1+x^2+x^3+...+x^55) because x_1 can take on values from 1 to 55 (inclusive) and so can x_2 and x_3. Since there are 3 terms in the sum x_1+x_2+x_3=111, that is why (x^1+x^2+x^3+...+x^55)(x^1+x^2+x^3+...+x^55)(x^1+x^2+x^3+...+x^55) has 3 factors..[each factor is (x^1+x^2+x^3+...+x^55)].

[errant bear]

that was the goal, when learning this topic it helps to write out your reasoning explicitly

[wise barn]

What is discrete math?

[cerulean radish]

Discrete is synonymous to finite

You mostly deal with finite or at least countably many objects

[cerulean radish]

Ah that makes a lot of sense

Thank you

[snow sleet]

Ur welcome!

[solar marsh]

wow how cool!! Thank you so much for sharing it.

[white rampart]

hi

[acoustic pond]

f:ℕ→ℝ with f(x)=3x-19

is this function injective, but not surjective?

[cerulean radish]

Well, what do you think?

[acoustic pond]

i think so because not every value in ℝ, in the range of value has a partner with ℕ

[brave rock]

And can you prove that?

[acoustic pond]

hmm I mean I could draw a function in a coordinate system and then it´s visible

or am I wrong

[latent geyser]

hello, does anyone have recommandations for practicing factorising (websites books?) I am having trouble with every math course in college since we didn't cover factorising well in highschool (p.s. any website that has exercices and tutorials for factorising all types of equations)

[brave rock]

I googled and found this website where you can practice factoring quadratics: https://www.mathsisfun.com/algebra/quadratic-factoring-practice.html

[cerulean radish]

Drawing the graph doesn't contribute a proof

For injectivity, assume that f(x) = f(y) for some natural x, y; Show that x = y follows

For disproving surjectivity, provide at least one real number excluded from the image of f (show why that is so too)

[pastel coral]

How many solutions $m \in \mathbb{N}\setminus {0}$ satisfy $140701 \overset{m}{=} 107041$?
$m$ would be a valid solution if gcd$(140701,m) | 107041$, is this correct? If no, why not?

If yes, I have another question

[brave rock]

You've overcomplicated this a bit

Just remember what the definition of equivalence modulo m is, it should be clearer

I'm not sure if the solution you've written here is right, just because I haven't checked if there's some silly business with the particular numbers, but this is not the general solution.

[pastel coral]

Oops one sec

wrong number

[vital dew]

cedric_

[pastel coral]

this the question

[brave rock]

I should say that I really am not worried about the particular numbers – I care only about the method.

[pastel coral]

but the remainder of LHS divided by m is equal to the remainder of RHS divided by m

but how to go from here?

[brave rock]

Remainders are not the easiest way to think about modular arithmetic.

Are you sure you've not seen a different definition?

[pastel coral]

oh, that's actually how we got taught them

[brave rock]

OK

[pastel coral]

maybe if you tell me ill remind it but for now i dont lol

ah wait maybe i know

[brave rock]

The definition that is the easiest is $a \equiv b \bmod m$ if and only if $m \mid a - b$. If you haven't seen this before, convince yourself that it's true first, and then it should be blatant what the answer is.

[vital dew]

Boyt

[pastel coral]

ahhh ye ok ofc, i know this yea

[brave rock]

Well then you should see the answer now

[pastel coral]

ahh ok, so this reduces to 140701-107041 = 33660, and now i just have to find divisors of 33660?

and ye and writing this in prime factorization will give us $2^2\cdot 3^2\cdot 5\cdot 11\cdot 17$

[vital dew]

cedric_

[pastel coral]

and from here we can count it out

[brave rock]

Well yes you could find the divisors I guess. The point is, it is precisely the divisors of 33660 that work.

[pastel coral]

should be 72

[brave rock]

Oh, you mean to say you count the divisors. Indeed 3^2 x 2^3 = 8 x 9 = 72, nice.

[pastel coral]

oh ye my bad

just dumb to forget about the simple definitions

but thanks a lot

[brave rock]

You see why thinking about remainders all the time is unhelpful, right?

This is a pattern, you should definitely keep it in mind

Remainders are good for computation, not for reasoning.

[pastel coral]

yup, will remember it now

[pastel coral]

ill keep that in mind, ty

@brave rock actually one more question

[pastel coral]

here i computed prime factorization with wolfram, but is there a systematic way to determine prime factorization by hand without any software?

or not really?

[brave rock]

Sure, keep finding prime factors and keep dividing lol

It is labour-intensive I'm afraid

Prime factorisation is a famously hard problem.

[pastel coral]

ye alright, i guess I wont have to expect such example on the exam then

thanks

[brave rock]

For small numbers (less than 200, probably) you may be expected to be able to compute prime factorisations.

But these aren't hard, usually.

[open island]

I am trying to solve this problem: https://projecteuler.net/problem=31

I have been trying to use a generating function to solve it. I have watched two youtube videos and read a section of a textbook regarding this problem type

video 1: https://www.youtube.com/watch?v=Lp2oEhGulL8
video 2: https://www.youtube.com/watch?v=3llKS5i7vV8

Here is my current work...
Let A be my generating function

$A = (1+x^1+x^2+x^3+...) \cdot (1+x^2+x^4+x^8+...) \cdot (1+x^5+x^{10}+x^{15}+...) \cdot (1+x^{10}+x^{20}+x^{30}+...) \cdot (1+x^{25}+x^{25}+x^{50}+...) \cdot (1+x^{50}+x^{100}+x^{150}+...) \cdot (1+x^{200}+x^{200}+x^{200}+...) $

that simplifies out to become

$A = \frac {1}{(1-x)(1-x^2)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100})*(1-x^{200})}$

In video 1 at 10:30 (click this link https://youtu.be/Lp2oEhGulL8?t=631, a taylor series expansion was done and explains how I can get my answer. The coefficient of $x^{200}$ in the taylor series expansion is what I want.

Taylor Series:
$f(0) + \frac {f'(0)}{1!} x^1 + \frac {f''(0)}{2!} x^2+...+ \frac {f^{(n)}(0)}{n!} x^n$

I want the value of this
$ \frac {A^{(200)}(0)}{200!} x^{200}$ where x is evaluated at 1

I am trying to find a method to find this value. I was hoping this would be the right channel. I believe that it involves combinatorics, specifically combinations but there does not seem to be a good resource out there.

[vital dew]

gloom

gloom

[icy swift]

just fyi: the easier you make it for people to answer your question the more likely it is someone will answer it

(thinking of the 3 links)

[pastel coral]

Let $V$ be a set that contains a followup of turns of a Rubiks Cube. Consider the operator $\circ = $ composition. Does $(V,\circ)$ for an Abelian group?

[vital dew]

cedric_

[pastel coral]

I'd say no, since there would be no identity element? Or could you state the identy element as like 4 times the same turn?

Also I suppose associativity isn't active here?

[fossil mural]

i'd say the sequence of no turns works as an identity element

[fossil mural]

...why wouldn't it be associative? do you have an example where associativity fails?

[pastel coral]

oh ye

[pastel coral]

one sec im visualizing it for myself first

ahh no indeed i see it doesn't fail, so it's should just be a normal non abelian group?

since commutativity fials

fails*

[little prism]

(be careful with the word normal cause it has a special meaning in group theory)

but yes

[pastel coral]

oh does it?

[pastel coral]

Question: is $$({ 2x | x \in \mathbb{Z} },+)$$ isomorphic with $$({ 2x -1 | x \in \mathbb{Z} },+)$$ I'd say no because the first structure is a group and the second is not a group, hence they cannot be isomorphic.

[vital dew]

cedric_

[pastel coral]

Is this right?

[cerulean radish]

If we are talking about group isomorphism, yeah

Groups can't be isomorphic to non-groups

[pastel coral]

ye, alright thanks

[brave rock]

A clarification on what bean said: it is completely meaningless to talk about isomorphisms between groups and non-groups — there is no definition for that. So yes, they are not isomorphic, but only because you may as well be talking about any other undefined concept.

Isomorphisms exist in other contexts, but this is not one of those contexts.

[devout ivy]

I am finding probability of a full house, this is a solution I got from discrete math 103 on study.com . Why did it use combination (4 2) ?

[stray reef]

presumably it's how many options there are for the rank of the pair

[devout ivy]

this is their a=explaination

[stray reef]

bc it can't be the same as the triplet

[devout ivy]

i still dont get like how is it (4 2)

hm

like how it say we need to consider the combination in the deck how is it 4 2

[stray reef]

4C2 is the number of ways to choose which suits go into the pair

also you're overusing the word "it" and that's probably contributing to your confusion

in order to construct a full-house hand we need to choose the following:

• what rank the trio is (13 ways)
• what suits go into the trio (4C3 ways, since we are choosing 3 suits from 4)
• what rank the pair is (12 ways, since we can't use the same rank as the trio)
• what suits go into the pair (4C2 ways, since we are choosing 2 suits from 4)

do you understand this @devout ivy Y/N

[devout ivy]

Y

do u help me clarify this

4 C 2 is to get something else to fill up our 5 deck?

along with the 12C 1

oh

wait

ithink its because im dumb

i did not understand what full house is in the first place

full house is 3 same card along with another 2 same card (different number)

i thought only 3 same card part matter, the 2 can be anything 😭

thank you so much Ann, thank you!!

[stray reef]

if the other two cards are different ranks then it is just called a 3 of a kind, not a full house

[prime grotto]

what is the difference between an ordered and an unordered increasing tree? i cant find the definition in the pdf im using

[iron steppe]

afaik (assuming binary tree) an ordered increasing tree is where each parent node has a left child that is less than itself, and a right child that is greater than itself, s.t. left child < parent < right child; an unordered increasing tree is where each parent node has children that are greater than it, but left/right doesn't matter, they just have to be >= the parent

[prime grotto]

oh thanks

[iron steppe]

np

[devout ivy]

this is last step in linear recurrence of homogenous

how is 5 = f1(3)^1 + f2(-1)^1 came out as f1 = 3/2 and f2 =-1/2

like how they obtained 3/2 and -1/2

[devout ivy]

can someone explain this for me, what does it mean by highest term

[stark spire]

does anyone have any youtube recommendations for graph theory?

[weary tiger]

books are way better

[livid wedge]

Have you tried khanacademy.org?

[stark spire]

no, but i will take a look

[stark spire]

i dont like books, i find it extremely difficult to absorb information

[cursive dew]

So, im wanting to get into the theory of math, and so far i heard words set theory and "axioms" but idk where to start at to go deep.

[devout ivy]

can someone double check my work, the answer is right but im not sure im heading the right direction

[stone lintel]

This seems like the right direction

[placid knot]

Can someone help me with this exercise of an old exam?

I don't see any of the questions being the right one, haha

In my opinion, it is not reflexive since not all vertexes has a loop. Not transitive because only {e,f,c} has transitivity. Not symmetric since only {a,b} are symmetric.

Perhaps the answer is antisymmetric?

[cerulean radish]

To disprove transitivity and symmetry, you need to provide a counterexample for each

Your answer is true, yeah, but you need to prove it

And, regarding antisymmetry, can you find any pair of different vertices x, y such that there is an edge from x to y and another edge from y to x?

[sterile atlas]

why isnt this transitive

can you find 2 pairs (x,y) and (y,z) such that (x,z) is not part of R

[cerulean radish]

"Why isn't this transitive?"
Proves that R is not transitive

[sterile atlas]

i did not proof that it isnt transitive

[cerulean radish]

Ah I misread

[sterile atlas]

i said that you cant find an example where transitivity doesnt hold

[cerulean radish]

My bad

[sterile atlas]

and it is surely not antsym bc (a,d) are in R and (d,a) are also in R but a doesnt equal d

(e,e) is not part of R so not reflexive

and also not symmetric cuz (e,c) is in R but (c,e) not

and ye for transitivity there is no counter example

which means that it is transitive

[pastel coral]

For c), how does this work?

The question is "Are these polynomials irreduceable or not? If they are not, give the factorization."?

[cerulean radish]

Clearly, there is no linear factor

See what happens when you assume x^4 + 1 is a product of quadratics

The coefficients are in Z_3 so the conclusion should arise pretty quickly

[pastel coral]

That's the part that I don't really understand, how should I understand this? Ofcourse for a) it's clear that i'll have two quadratics, and for b) i can reduce them more to the complex solutions, but how does it work with Z_3?

[pastel coral]

(x^2-i)(x^2+i), so this can't be reduceable since it's complex?

[cerulean radish]

Assume there exist $a, b, c, d \in \bZ_3$ with $x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$, try to solve for them or prove that there is no solution for them

As a side note, we can assume that the coefficients of x^2 in those quadratics can be 1 because if they were 2, we would just multiply both factors by -1

[vital dew]

A Lonely Bean

[cerulean radish]

The polynomial being irreducible in R does not mean it is irreducible in Zn

[pastel coral]

ahhh

[cerulean radish]

E.g., x^2 + 1 is irreducible in R and is irreducible in Z2

That statement would be fine if Zn was a subfield of R, but it is not

They have different arithmetic after all

[pastel coral]

One sec, i'm analyzing this here

ill be back

[pastel coral]

there exist no solutions in Z3, so it's irreduceable?

[ruby iris]

if we have statements a,b,c
my prof said that if a -> b, b->c and c->a, then a,b,c are all equivalent. But why dont we also need to show c->b????

[cerulean radish]

But there is a solution

I think I had found it

Let me recheck

[cerulean radish]

Yeah there is a solution

[pastel coral]

Huh but not in Z3 right?

[cerulean radish]

In Z3

[pastel coral]

huh

or is my system of equations wrong?

[cerulean radish]

The system that you should have is a + c = 0, b + d + ac = 0 and bd = 1

[pastel coral]

huh, you need to have 4 equations no?

[cerulean radish]

Where does ad + bc = 0 come from?

[pastel coral]

for constant term, term in x, in x^2 and x^3?

[pastel coral]

euuh one sec, i removed my notes lol

just from these

[cerulean radish]

Ah you are right

[cerulean radish]

Wolfram is trying to solve in Z

Not in Z3

[pastel coral]

ah, but how do we solve in Z3 then?

that's what i'm confused on

[cerulean radish]

c = -a, so b + d = a^2

Since bd = 1, you have two cases

b = d = 1 or b = d = 2

Continue with this

[pastel coral]

ohhh yeah cuz bd = 4 but modulo 3 it is 1

I think i can go from here on, thanks a lot

@cerulean radish

(x^2+2x+2)(x^2+x+2)

this is it right?

[cerulean radish]

Yes

[pastel coral]

ahhh awesome, thanks for the help

i just didnt knew what to understand to reduce it over Z3

got it now

[autumn perch]

i am trying to solve this set question. But my answer is not correct. How can i solve the question (b) and is the answer of (a) correct?

[weary tiger]

hey someone help me w chinese reminder theorem pls?

[lofty cloud]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[gusty swallow]

your answer to a is 450?

[modest swan]

what exactly is the difference between propositional and predicate logic

[frozen mason]

I’m trying to prove that the subset of the linear code C whose members are the code words with even weight is also a linear code

I’m out of ideas

The goal was to show that for a and b inside this subset call it E d(a,b) = 2k for some integer k and then it follows that d(a+b,0) = 2k as needed

[haughty garden]

Is C a linear code over Z_2? If so, then look at the places where a and b agree; these become 0 in a + b

say that a and b agree on k coordinates, then d(a + b, 0) = (d(a, 0) - k) + (d(b, 0) - k) = d(a, 0) + d(b, 0) - 2k, which is even because a and b are in E

[frozen mason]

is this some identity?

[haughty garden]

Throughout, I will assume that we are working with binary codes. You can prove it using some set theory. Let $a = (a_1, \dots, a_n)$ and $b = (b_1, \dots, b_n)$ be two codewords and suppose that $a$ and $b$ agree on $k$ coordinates, that is, $a_{i_j} = b_{i_j}$ for $j \in {1, \dots, k}$. Also, let $I_a$ denote the set of non-zero coordinates of $a$ and let $I_b$ denote the set of non-zero coordinates of $b$.

Firstly, convince yourself that the coordinate $(a + b)_i$ is nonzero iff $a_i$ and $b_i$ differ (i.e. either $a_i$ is 1 or $b_i$ is 1 but not both). This implies that (or convince yourself that) [d(a + b, 0) = \lvert I_a \triangle I_b \rvert = n - k.]
Remember that the symmetric difference is the set where a coordinate is nonzero in either $a$ or $b$ but not both! Now, with some basic set theory, we see that
\begin{align*}
d(a, 0) + d(b, 0) &= \lvert I_a \rvert + \lvert I_b \rvert \
&= \lvert I_a \triangle I_b \rvert + 2 \lvert I_a \cap I_b \rvert \
&= d(a + b, 0) + 2\lvert I_a \cap I_b \rvert.
\end{align*}

In fact, it's enough to stop here but see if you can convince yourself what $\lvert I_a \cap I_b \rvert$ represents (i.e. it is the set of coordinates on which $a$ and $b$ are both nonzero). Think about why this identity is true intuitively.

[autumn perch]

it is 20 after recalculating. it is the

| F intersect B intersect H| = | F U H U B| - |F| -|B| - |H| + | F intersect B| + | F intersect H | + | H intersect B|
= 450 - 285 -115 - 195 + 45 + 70 + 50
= 20

is this correct?

[fringe turret]

Based on this https://fair-use.org/bertrand-russell/the-principles-of-mathematics/s37 could someone help me understand the meaning of formal implication and how it is different from material implication? I understand that it is called material because of relation between two proposition, but what does formal implication mean?

The language used here is difficult for me to comprehend.

[placid knot]

Hi. Would anyone be able to help me with some proofs in language grammar?

[lofty cloud]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[coral parcel]

Propositional logic doesn't have quantifiers. Formulas are built from AND, OR, NOT, IMPLIES, and "propositional variables" whose values are either "true" or "false".
Predicate logic extends this with variables whose values can be more interesting objects. In addition to AND, OR, NOT, IMPLIES, there are quantifiers for binding the variables, and instead of propositional variables, an "atomic formula" consists of applying a predicate to one or more variables (or expressions built from variables and functions).

[placid knot]

Can anyone explain this? I don’t get what the answer is. Super hard. Been trying for 2 hours

[errant bear]

P(A and B) = P(A) * P(B)

[mild temple]

Events A and B need to be independent for this to hold.

[mild temple]

!status

[lofty cloud]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mild temple]

oh wrong command. it shld b this one

[lofty cloud]

Show your work, and if possible, explain where you are stuck.

[modest swan]

Thank u very much

[fringe turret]

Could someone help me with this? I haven't learnt about formal implication before this. All I know is about formal logic.

[brave rock]

This is a purely philosophical distinction that has little to no effect on how one does mathematics.

Russell seems to be saying that although it is 'formally' true that, for example "1 + 1 = 2" implies "The Eiffel Tower is in France" since they are both true statements, this doesn't fit with how we really think about how things imply each other. There is material implication such as "I am on the eiffel tower" and "the eiffel tower is in France" allows me to conclude "I am in France." He does not define these explicitly, because once again this is philosophy and not mathematics.

[fringe turret]

I see, maybe thats why I didn't understand 😅

Thanks for your response

[pastel coral]

Can someone explain me the symmetry group of a square prism? If I have to believe Wikipedia the order of the symmetry group should be 16.
Although I got 12 due to the following thinking:

• 5 symmetry planes (XY, XZ and YZ planes and 2 planes like the diagonals of the square)
• 5 rotations (one rotation of 180° "around" each plane, idk how to explain)
• the identical rotation
• point reflection

Where am I wrong?

[odd heart]

You've got 8 symmetries corresponding to the symmetries of the square, and in addition you can compose each of them with the reflection in the symmetry plane that's parallel to the square and divides the prism in the middle.

That gives you 16 symmetries, and you still need to show that these are all of them, but it's probably not hard

[pastel coral]

Ah, that makes sense, thanks a lot

For basically for ANY cuboid with depth != height != width the order would be 8, 4 of the rectangle and then reflection in the symmetry plane is 2x4=8

[odd heart]

Seems that way

And the cube has loads of symmetries and different stuff happpens there entirely

[pastel coral]

ahh, i see, thanks a lot

[plain ruin]

Can anyone decipher what my professsor has written on the board where I highlighted with question-marks? I don't understand how he simplified it from the previous stage to that.

[little prism]

well m^2+4(m+1)=m^2+4m+4=(m+2)^2

and the rest is basically just carried along

[stone lintel]

$$\begin{align}
&=\dfrac{m^2(m+1)^2}{4}+(m+1)^3\
&=\dfrac{m^2}{4}(m+1)^2+(m+1)^3\
&=(m+1)^2(\dfrac{m^2}{4}+(m+1))\
&= \dfrac{(m+1)^2}{4}(m^2+4(m+1))
\end{align}$$

[vital dew]

catman
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[plain ruin]

Thanks! I solved it (:

[brave rock]

You should delete that offer. From the rules:

Do not offer money for doing homework assignments, and vice versa.

Also

!da2a

[lofty cloud]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[placid knot]

Can anyone tell me if I chose the correct answer?

[livid wedge]

How did you get there?

[hard citrus]

is there anyone here who speaks hebrew to help me with some questions i have about proofs on set theory?

[hard citrus]

can someone review my proofs in set theory and tell me if im missing something/done some errors?
im a first year and its my first course in discrete math and proofs, and the prof isnt very good at explaining.

[frozen mason]

What is the triangle symbol I havent seen this before

[cerulean radish]

The symmetric difference

[frozen mason]

Ok what's that?

[cerulean radish]

You basically take all the elements that are either in A and not in B or in B and not in A

[frozen mason]

Ok

[cerulean radish]

I.e., it's the union of A and B minus their intersection

[frozen mason]

So all the elements that is not in the intersection?

Ok yeah

Thanks.

[quiet eagle]

Consider the set $S = {1, \dots, n}$, where $0 \leq k < \tfrac{n}{ 2}$
. \Let A and B represent the sets of all subsets of $S$ with $k$ and $k+1$ elements, respectively. \
For $C \subseteq A$, we define $N(C) \coloneqq { b \in B \mid \exists a \in A \colon a \subset b }$.\
\ Show that: $\forall C \subseteq A$, $|A| \leq |N(A)|$.

[vital dew]

Sciencenjoyer

[quiet eagle]

Hi, I need a little hint for this. My only idea is that, due to the choice of k, nCr(n,k) < nCr(n,k+1) (binomial coefficients).
So at least A is smaller than B. (sry for repost)

[cursive dew]

Just bumping, but again i dont really know where to begin. Trying to find the "why" of math.
I assume axiom stuff is the reason why a negative times a negative turns it into a positive?

[brave rock]

You can derive that from the ring axioms, yes. This is to say that if we require a list of reasonable properties of addition and multiplication, such as a(b + c) = ab + ac, then this necessitates (-a)(-b) = ab.

[cerulean wind]

check your latex. the thing u intend to show doesn’t depend on C at all

[quiet eagle]

ohh, thanks for telling

Consider the set $S = {1, \dots, n}$, where $0 \leq k < \tfrac{n}{ 2}$
. \Let A and B represent the sets of all subsets of $S$ with $k$ and $k+1$ elements, respectively. \
For $C \subseteq A$, we define $N(C) \coloneqq { b \in B \mid \exists a \in C \colon a \subset b }$.\
\ Show that: $\forall C \subseteq A$, $|C| \leq |N(C)|$.

[cerulean wind]

N(C) is defined incorrectly as well but now it’s more clear

[quiet eagle]

Is it? Whats wrong there?

[cerulean wind]

doesn’t depend on C at all

[quiet eagle]

ops

[vital dew]

Sciencenjoyer

[cerulean wind]

for each set in C, append the smallest element of S which is not in the set. this is a function from C to N(C)

what properties does this function have?

[quiet eagle]

It seems to be neither injective nor surjective.. I'll look for something more useful

[cerulean wind]

i believe it’s injective

[sick grail]

it is not

e.g. both {1} and {2} (for k=1 when n allows that) map to {1,2}

[quiet eagle]

If this function is called f, we certainly have |C| >= |f(C)|

[quiet eagle]

I got no progress :(
Any ideas?

[haughty garden]

It looks like you could probably apply Hall's theorem

Form a bipartite graph with A on one side and B on the other side, where a set A' in A is connected to a set B' in B iff A' is a proper subset of B'

[quiet eagle]

The only relevant property of bipartite graphs I know is Königs theorem. So I'll try that, thanks

[haughty garden]

Yeah, Hall's Marriage Theorem should work here

[quiet eagle]

Is the marriage theorem a special case?

We generally call it Hall condition for an A-perfect matching

[haughty garden]

Hall's theorem can be proven with Konig's theorem

I think Diestel proves Hall's theorem using Konig

going back to the original problem, you need to show that the bipartite construction satisfies Hall's condition and doing so immediately implies the result

[quiet eagle]

I see. I am a bit afraid of the property of the subset relation between these sets, which I'll probably need to use

But I'll see

@haughty garden Are you sure, you ment the marriage theorem as it needs |A| = |B|, which we don't have?

[haughty garden]

you don't need |A| = |B|

the graph formulation of Hall's theorem states that a bipartite graph has an A-perfect matching iff, for each subset C of A, |C| <= |N(C)|

[quiet eagle]

I agree

[haughty garden]

so each element of A needs to be matched to a unique element of B

but B need not have a perfect matching

[quiet eagle]

In my lecture, the "marriage theorem" is a special case of Halls Theorem, so I was confused

But we're talking about the same condition

[haughty garden]

ok I'll just refer to it as Hall's theorem

I interchange between the two since I've always been taught them as the same

just cbf saying marriage half the time lol

[quiet eagle]

Oh ok lol

[haughty garden]

actually you probably don't even need it

can just prove it directly

[haughty garden]

construct the bipartite graph ^

it's firstly not hard to show that for any set P in A, deg(P) = n - k because you can just extend P to a (k + 1)-element set by appending some element you haven't used yet; there are n - k such elements to choose. Also, any set Q in A has deg(Q) = k + 1.

Take any arbitrary subset C of A, and let E be the set of edges from C. Similarly, let E' be the set of edges from N(C). Each edge in E belongs in E', so E \subseteq E' and so, |E| <= |E'|.

Now, the number of edges in E is given by |E| = |C| * deg(P) = |C| * (n - k) and |E'| = |N(C)| * deg(Q) = |N(C)| * k + 1, so ```
|N(C)| = |E'| / (k + 1)

= |E| / (k + 1)
= |C| * (n - k) / (k + 1)
= |C|.

[quiet eagle]

I read the first lines. I'll close my eyes now. Thank you for giving me the direction

[quiet eagle]

I got it, ty

[elder berry]

@haughty garden is Diestel the way to go if I wanna start advanced graph theory?

I know the basics of graph theory, pretty much the first chapter of Diestel, do I continue through the rest of the chapters as well or would you recommend smth else?

[haughty garden]

I would recommend sticking with Diestel if you want to get the full flavour of graph theory, but I suppose it depends on what you’re aiming to achieve by the end. Another awesome reference book is Bondy and Murty’s text which is possibly more suited to those who want to learn graph theory for computer science. Bondy and Murty covers the graph theory concepts that you typically find in a more advanced theoretical computer science setting, such as matchings, colourings (chromatic number, edge colourings, vertex colourings), independent sets and cliques, etc

[elder berry]

yeah aiming to study graph theory for cs, so definitely gonna check that book out too, ty!

[haughty garden]

Yeah okay, Bondy and Murty might be better for you then. They discuss some computational complexity stuff throughout so it ties nicely with the stuff you’d be familiar with in a standard computer science curriculum

Diestel is more suited if you were aiming to do pure math research in graph theory, whereas B+M is very much an applied textbook on graph theory

[elder berry]

interesting

I think I'll start with B+M and continue with Diestel, because I am kind of yearning for pure math, which my cs major fails to deliver haha

[haughty garden]

sounds like a good plan then!

[stone lintel]

@elder berry if you are looking for some more theoretical aspects in cs I would recommend the book "introduction to algorithms"

[elder berry]

will check it out! thanks

[brave rock]

You need to be a little more precise with that, but in general no you cannot determine the cardinality of the quotient from the information taht all the equivalence classes have a given cardinality.

E.g. Z/2Z is finite and the equivalence classes are countable, but Z with the equivalence relation generated by x related to 2x is different.

[frail scarab]

E={x=a+b√3/ (a;b)€Z×Z}
We define on E the relation R
xRx' <==> a-a' and b-b' are even numbers
After showing that R is an equivalence relation
How do i determine card(E/R)

[brave rock]

You think about it and try things

Hint: you can just think about Z x Z

[empty condor]

Yall got any sheets that might be helpful for proof by mathematical induction, contradiction or contraposition

It would be rlly helpful

[spring hound]

Do you know why all those work?

[spring hound]

Mathematical induction is like a machine that you make. The machine is where you prove that if it works for one number, it also works for the next. Then, you prove it for a specific number, like 1.

Now you can use your machine to say that since it works for 1, it works for 2. Then you can use your machine to say that since it works for 2, it works for 3. Then you can use your machine to say that since it works for 3, it works for 4. For any integer you choose that's at least 1, you can use your machine a certain number of times to go from 1 to 2 to 3 and so forth until you reach your chosen integer. So, the statement must be true of all integers that are at least 1.

---

Proof by contradiction is based on the idea that I say something, then I prove that what I said would force a contradiction (A and not A) to happen. Since contradictions are impossible and what I said has to cause a contradiction, what I said must be impossible, so some alternative to what I said must be true instead.

If you have a statement that's either true or false and those are the only two alternatives, then if one of them is impossible, the other must be the case. If there are more than two alternatives, proving that one of them leads to a contradiction only knocks that one out of the running.

---

Proof by contraposition is based on the idea that A -> B means that the combination of A being true and B being false shouldn't happen. Not B -> not A means that the combination of not B being true (B being false) and not A being false (A being true) shouldn't happen. So, since they both say that that A being true and B being false shouldn't happen, they're perfectly equivalent. If you prove one of them, you prove the other and if you disprove one of them, you disprove the other.

As a mnemonic, you can think of "contra" as putting nots on both parts (A -> B becomes not A -> not B) and "position" as switching their positions (not A -> not B becomes not B -> not A).

[empty condor]

Oh great

Damn

[empty condor]

Yes

I just want some practice questions yk

Got a final in a couple of days

[spring hound]

Here are some induction questions: https://www.math.waikato.ac.nz/~hawthorn/MATH102/InductionProblems.pdf.

Some proof by contradiction questions (at the end): https://cgm.cs.mcgill.ca/~godfried/teaching/dm-reading-assignments/Contradiction-Proofs.pdf.

[empty condor]

Greattt, thanks a lot, got any idea if they got the answers too?

[spring hound]

I think they don't on that one.

Some proof by contrapositive questions (at the end): https://www.people.vcu.edu/~rhammack/BookOfProof2/Contrapositive.pdf.

[balmy minnow]

what is the collatz conjucture?

i need some knowledge about it.

[fringe turret]

In this shouldnt the implication goes from A to B, because if is associated with A?

A means Albert stole cookies and C means Clive stole the cookies from the jar

[fossil mural]

start at some nonnegative integer, like 7
at each step, if it's odd, multiply by three and add one, and if it's even, divide by two
so in the case of 7, you get 7 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 4 -> 2 -> 1
once you get to 1, it goes back to 4 so you just end up in a loop
the collatz conjecture is the statement that whatever number you start at, you will always eventually reach 1

it's a "conjecture" because, while it's probably true, nobody has been able to actually prove it

[balmy minnow]

hmm thankyou

[fringe turret]

In the problem, I am struggling with rationale for why we have used the conjuction at last to get the details of who actually stole cookies from jar?

[fringe turret]

Ok this is clear now. I checked there are more aliases to this used in the natural language, if p, then q is same as p only if q. I confused it with iff.

[meager prawn]

Find a nice system of representants

And find its cardinak

what do you know about set quotients

[meager prawn]

i.e. explicit what E/R is

its elements are the classes

Z/2Z is very finite

That's the group quotient notation
Basically a ~ b iff a-b in nZ
With E = Z
Then E/R is called Z/nZ

Hence Z/2Z has 2 elements so it's not infinite

[meager prawn]

I didn't claim it to be related

I just gave a counter-example to this

[meager prawn]

It is, in fact, related
But that's for you to prove

Do you know the concept of representant of a class ?

[ionic lintel]

A chess board of size 6×6 can be covered by 18 domino stones of size 2×1.
Prove that for each such covering there exists a straight line (vertical or horizontal) that cuts the board without cutting any domino stones.

Help?

[sick grail]

Show that any line cuts through an even number of pieces, and use that fact in a proof by contradiction

[rough totem]

Hello I would like an honest opinion from someone who has taken this class before, how much time should you put into a week for this class?

[ionic lintel]

Thanks.

[hollow yarrow]

Is this the Union of A with itself n times and then infinite times?

[errant bear]

no

A_j isn't necessarily A_k

[fossil mural]

what's "this class"?

[fossil mural]

it's the union of n sets

so for instance if n = 3, then $A = A_1 \cup A_2 \cup A_3$

[vital dew]

bee [it/its]

[fossil mural]

where A_1, A_2, A_3, are just arbitrary sets, we're given that they're countable but we don't know anything else about what they are

[cerulean wind]

they mean discrete math, the name of the channel

[fossil mural]

well discrete maths is an area of maths, not a class

if he's planning to take a class in which "discrete maths" is the subject, then unless he's going to take all such classes in existence simultaneously, he needs to clarify which class he's considering taking

[cerulean wind]

it’s a course title at some uni’s

[little prism]

its a very unspecific course title

[inland nacelle]

Some weeks ago I had an exam and I had to prove that the only solution of this equation 5^x+2=7^y was x=y=1. I couldn't find a solution anywhere, I just had as a hint that using mod 25 was useful, but I couldn't figure out how to do it. Can somebody help me with it?

[little prism]

what can you tell me about 5^x mod 25

[inland nacelle]

well, 5^x its 0 in mod 25 if x>=2

so 7^y = 2 mod 25

i tried to work with the last digit of 7^y but that way didn't work

[little prism]

what are the first few values of 7^y mod 25

[inland nacelle]

7, 24, 18, 1

[little prism]

and then it repeats

so you never get 2

[inland nacelle]

ohhh

thank youuu

[little prism]

its easier btw to replace 24 by -1

[inland nacelle]

yeah

[hollow yarrow]

I’ve just watched a video on how to create a bijection from the naturals to the rationals to prove the rationals are countable with that snake thingy

So 1 -> 1/1, 2 -> 2/1, 3 -> 1/2, 4 -> 1/3, 5 -> 2/3, 6 -> 3/3 and so on

Is there a formal way to describe this bijective function?

I do understand the visualisation to it and why it proves the countability of Q

[brave rock]

There is a nice polynomial that provides a bijection N → N^2 — maybe you can try and find it, in fact! — and then Q can be embedded as a subset of N^2, hence as a subset of N. Then you use the usual trick on infinite subsets of N. This is the "formal" way to prove it.

It is not at all easy to write this function down, and I'm not aware of a nice form for it. In general one cannot write down bijections explicitly even if they are guaranteed to exist.

[calm sapphire]

im taking my first discrete math course, any tips ?

[foggy sentinel]

Before i head into my math test tomorrow, i get a few hours to spare. If you were me, what would you do ?

@harsh frost ;-;

[little prism]

sleep

[rough totem]

discrete structures

I believe this chat is for it

not sure

[rough totem]

I’m also asking

[harsh frost]

Yea probably would second the “sleep” one, mind you not super sure what I would recommend really
Probably of course something like “read over stuff” I guess, but generally what I’ve done depended on the topic and how much I thought I could do with it

[foggy sentinel]

as long as i get 9h of sleep it's fine for me :P

However i do have a question :s

6^8 mod 13

What is a simple way to solve this considering if i use fermats little thereom i will result in: 6^12 = 1 mod 13
And that isn't really something i can utilize afaik.

[prisma jolt]

What do for some and for all mean?

It seems like theyre both rhe same formulas except the for some and all parts, but i cant figure out what they mean

[supple talon]

for some means there is at least one such thing

for all means that it is true of all such objects

[prisma jolt]

At least one such what in that situation? Since that would read as at least one k, and that doesnt make sense i dont think?

[supple talon]

yes, there is at least one k such that x\in A_k

where 1\leq k \leq n

[prisma jolt]

thank u so much

figured it out now

[noble mango]

can you help me solve some practice?

[acoustic pond]

((b → a) ↔ c) =
((b → a) → c) ∧ (c → (b → a)) =
(¬(¬b ∨ a) ∨ c) ∧ (¬c ∨ (¬b ∨ a)) =
((b ∧ ¬a) ∨ c) ∧ (¬c ∨ ¬b ∨ a) =
(b ∧ ¬a) ∨ c∧ ¬c ∨ (¬b ∨ a) =
(b ∧ ¬a) ∨ 0 ∨ (¬b ∨ a) =
(b ∧ ¬a) ∨ (¬b ∨ a) =
(b ∧ ¬a) ∨ ¬b ∨ a =
¬b ∨ ¬a ∨ a =
¬b ∨ True = True

hi, where is my mistake?

[heavy nova]

I think something happened between lines 4 and 6, what are your justifications for those steps?

[heavy nova]

Something happened going from line 4 to line 6, when you proceeded to combine c ⋀ ￢c

I think you were assuming associativity holds interchangeably between ⋀ and ⋁, which it doesn’t. In fact, in P ⋀ (Q ⋁ R) and P ⋁ (Q ⋀ R), distribution holds but not association

[acoustic pond]

does this look better?

((b → a) ↔ c) =
((b → a) → c) ∧ (c → (b → a)) =
(¬(¬b ∨ a) ∨ c) ∧ (¬c ∨ (¬b ∨ a)) =
((b ∧ ¬a) ∨ c) ∧ (¬c ∨ ¬b ∨ a) =
(b ∧ ¬a) ∨ c ∧ ¬c ∨ (¬b ∨ a) =
(b ∧ ¬a) ∨ 0 ∨ (¬b ∨ a) =
(¬a ∧ b) v a v ¬b v 0 =
a v b v ¬b v 0 =
a v 1 v 0 =
a v True v False =
a v True =
True

[fossil mural]

(b ∧ ¬a) ∨ c ∧ ¬c ∨ (¬b ∨ a) it's not clear what this means, where exactly do the brackets go?

[acoustic pond]

think here it´s faulty, I think I need to stay with the brackets

(c ∨ b) ∧ (c ∨ ¬a) ∧ (¬c ∨ ¬b ∨ a)

i think further I can´t convert/transform it

[severe path]

Hello, I have an algorithm and I am trying to express the average time complexity in closed form

The average time complexity is of the form

$$\frac{\sum_{i=1}^{n}\sum_{j=1}^{n+1} |n-i-j+2|\binom{n+i-j}{i-1}\binom{n-i+j-1}{j-1}}{\binom{2n}{n}}$$

I have computationally arrived at an approximation of $\frac{4}{9}n^{\frac{3}{2}}$, however I want a more rigorous proof of how to get there. Aside from trying to substitute stirlings approximation, which I have tried, is there entry into this problem?

[vital dew]

QuAnTuM

[fringe turret]

In the answer key of 5th part, it is $p \wedge q \to \neg r$, why not $(p \wedge q) \to \neg r$?

[vital dew]

ĐARK々MÁTTER

[fringe turret]

Also please could someone help me with this? TA of the course has ignored it :/

[weary tiger]

“If at least one of them stole it” and “if Clive only steals it Albert steals” and “if Betty doesn’t steal alone”, then …

I’d assume that why they used conjunctions

Basically, they used conjunctions to make it such that all these statements are true at the same time

[icy swift]

was told to put this here so im just copy and pasting it

I have a question for a beginner problem in my combinatorics textbook
"How many odd numbers between 1000 and 9999 have distinct digits?"

So if we make 4 placeholders for the thousands, hundreds, tens, and ones place. We can see that if we start with the ones, there are 5 possible choices, then check for the thousands there are 8, then 8 for the hundreds, and then 7 for the tens, so the answer is 2240
But if we instead go from saying there's 5 possible choices in the ones, and then go to the tens, there would be 9, and then 8 for the hundreds, and then 6 for the thousands, which would give 2160

[haughty garden]

the issue with the second method of counting is that when you place a 0 in either the tens or the hundreds place, you no longer have 6 choices for the thousands place

e.g. if your number is _ 1 0 9, how many ways can you fill in the first digit?

[icy swift]

ohh thats right

[icy swift]

So you cant use multiplication principle if you start with the tens after the ones place since the number of choices for the thousands place changes whether or not a 0 already exists in the tens or hundreds

[haughty garden]

it's more like you didn't count it correctly

if you want to use the second method of counting, you need to break it into two cases: one where the 0 has appeared in either the tens or hundreds place, and one where the 0 doesn't

but that's more work than just doing it the first way

[icy swift]

i see

makes sense now thanks for the explanation

[haughty garden]

👍

[rich ledge]

Let's give you all all a fun exercise\
\newline
Given the relation $R$ defined as :-
\begin{equation*}
R = {(a,b) \in Z \times Z | even(3a - 5b)}
\end{equation*}
Prove that R is an equivalence relation.

I saw this in my discrete final exam, man it was a shock to see that as the first question test. Though it was fun to solve, yall should try your luck at it

[vital dew]

tooBoard

[fossil mural]

solution ||aRb iff a = b mod 2, QED. now here's some padding text so that the length of the spoiler tag doesn't give away how short the solution is||

[fringe turret]

There are 2 propositions
p: It is below freezing
q: It is snowing.

I want to write the symbolic form of: Either it is below freezing or it is snowing, but it is not snowing if it is freezing.

This is what I came up with: $(p \oplus q) \wedge (p \to \neg q)$, but in the answer key it is $(p \vee q) \wedge (p \to \neg q)$.

[vital dew]

ĐARK々MÁTTER

[fringe turret]

Second part it clear to me, but in the first shouldn't it will be exclusive or, because of Either...or...?

[vestal bronze]

exclusive or would be sufficient on its own

so they are both right, but the book answer kinda avoids overtly "wasting space", i guess

[fringe turret]

$\vee$ means inclusive or and $\oplus$ means exclusive or, their truth tables are different. How come they both be right (same thing)? Also, I didn't understand "exclusive or would be sufficient on its own", please help me understand it.

[vital dew]

ĐARK々MÁTTER

[vestal bronze]

the truth table for the entire statement is not different, that's why i said they are both right

and p ⊕ q has that truth table too

so all three are correct

if the answer key uses ⊕ in every other situation like this, I would agree it's wrong, there would be a contradiction

without that context, it just gives one of the right answers

[fringe turret]

Wait what?

When we are given P and Q are sufficient for R, then it is same as $R \to (P \wedge Q)$, right? or the other way around?

[vital dew]

ĐARK々MÁTTER

[fossil mural]

$p \oplus q$ and $p \land q$ don't have the same truth table \ $(p \oplus q) \land (p \to \neg q)$ and $(p \lor q) \land (p \to \neg q)$ do

[vital dew]

bee [it/its]

[fringe turret]

I meant XOR and OR operator.

[fossil mural]

the case where p and q are both true, which is the only point where they differ, it doesn't matter, because then p -> ¬q is false, so the entire expression is false either way

[fringe turret]

Btw, after his frownyfrog's message I validated no matter what value of P and Q in this particular case we can use either xor or or. But, because of the Either...Or thing, I chose XOR, thats now my default for me

[weary tiger]

Hi I can't understand why for non homogeneous part instead of C2^n this is replaced by Cn2^n? According to guess table we will guess r^n as constant multiple of r^n for non homogeneous part 2^n but it is not the case here.

The solution will be of form (A+C)2^n and I see nothing wrong here...

but it is 2^n(1+n) instead

May be you can ignore me. It seems this is shut up and compute stuff.

[quaint flower]

Question, do you describe permutations as:
Choices, where the order of the elements matter? I'm not sure if I can describe them properly. Anyone who knows a good explanation?

[cerulean radish]

Permutations are usually defined as bijections from a set into itself, mostly in the context of finite sets

[quaint flower]

If we exclude set theory then? Can you explain this concept in a more... non-set theory matter?

It was a long time ago since I worked with bijections. But, isn't that a mapping that is injective and surejective?

are you simply implying that you can map set A from set B in such a way that:

All elements from A and B are the same? Therefore, making a direct 1-1 mapping between them.

[cerulean radish]

Yeah, a mapping is a bijection that's injective and surjective; In this context it will implement the way the objects are permuted from their initial state to whatever current order they have

[cerulean radish]

The domain and the codomain of a mapping can be the same sets, yes

[quaint flower]

So if the domain is the same as the codomain, for example:

Domain: A(set)
Codomain: B(set)

Then you'll have this:

A = {x | x in B}
B = {x | x in A}

Right?

[cerulean radish]

Sure

Or just write A = B

[quaint flower]

Oh sure

[quaint flower]

Like, if I have 3 people, and I'm supposed to sort them into 3 groups.

I'll pick 1 person at a time.

You'd have 3! permutations right?

3x2x1

[cerulean radish]

If the order of the groups mattered, yes

[quaint flower]

Do you mean like, by that, AxBxC

Is not the same as CxBxA

[cerulean radish]

Yeah

[quaint flower]

In some theoretical situation

ah

[cerulean radish]

Where A, B and C are the groups

[quaint flower]

Well, if I picked 2 options at a time.

Then you have something different right?

So, I'd have: 3!/(1!) permutations?

or 2! at the denominator

[cerulean radish]

Wdym by picking 2 options at a time?

[quaint flower]

Like, you pick a pair of people and sort them into groups AxBxC

If let's say, you had 3 people and paired them

I'm not sure how to explain the question

because my terminology sucks

[quaint flower]

I mean: (Let's take a new example)

If you had three groups:

A
B
C

And you had a total of 6 people.

How many permutations are produced if you pair 2 people into each group?

Or does it not work, because you have to specify which kinds of people?

[cerulean radish]

Ah

Hmm

So for group A you have 6 * 5/2 choices
Once group A has been chosen, group B has 4 * 3/2 choices

And once groups A and groups B have been chosen, group C has 2 * 1/2 choices

So It's 6!/2^3

,w 6!/8

[vital dew]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by alonelybean

[quaint flower]

i'll have to study this more intently before I understand this though

Because erm, me brain stoopid.

I have recently grasped some combinatrorics related to multiplication and addition principles.

[severe path]

I found this identity on stack exchange, can someone explain where that integral definition comes from? https://math.stackexchange.com/questions/2008583/closed-form-for-weighted-sum-involving-product-of-binomial-coefficients

[severe path]

nvm its a fourier trick

[quaint flower]

Erm, how do you know the amount of permutations on a 4 digit lock?

I'm not really sure about the process... since, I find permutations hard to understand

[pale monolith]

what do you mean by permutations on a 4 digit lock?

usually permutations is about the number of ways you can place something in order

[quaint flower]

I can't frame the questions properly

Because, errr, it's a new concept and I am having serious trouble even grasping it.

I have grasped addition-multiplication principle, but yeah, gah

[pale monolith]

is it about the number of different possible ways to lock the lock?

[quaint flower]

yes

[pale monolith]

ok

we'd use the word combination for that

maybe that's why it's called a combination lock

[quaint flower]

What about the order mattering for the total amount of ways to lock it? Then it would be permutations right?

duplicates cannot exist in permutations, right? or, they do exist

[quaint flower]

Yeah but I am doing permutations

not combinations

wouldn't that lock produce 10^4 combinations? Or is it 10^4 permutations? I have no idea.

I'd agree there are 10^4 ways to do it, but, I am just lost at this point

[pale monolith]

yeah it's 10^4

for each digit you have 10 possible options

[quaint flower]

Well yes

[pale monolith]

the order does matter in a lock

1001 is not the same as 0110

[quaint flower]

10 choices
10 choices
10 choices
10 choices

^ One digit and the second digit and the third digit and the fourth digit

10^4

[pale monolith]

exactly

[quaint flower]

I know that part, but then when it comes to permutations, it becomes... confusing

should I say 10^4 permutations? Well, I'm getting 10!/(6!) permutations instead

[pale monolith]

a permutation of some group of objects is just a specific way to order them

so ACB is a permutation of ABC

[quaint flower]

sure, I can buy that.

I know there's at least 3! permutations on that

[pale monolith]

there are in fact exactly 3! permutations

[quaint flower]

if we chose 1 element from 3 elements

yes

If we however, chose a pair of elements from these 3! permutations, how many ways can you do that?

3!/(2!)

Three permutations

[quaint flower]

But the problem is, is it the amount of permutations you chose from? Or the amount of elements?

I'm not sure how to describe it

[pale monolith]

there are 3!*(3!-1)/2 ways to pick two permutations from the permutations of ABC

[quaint flower]

So not three?

[pale monolith]

yes, not three

[quaint flower]

...I don't understand why

There is a formula for that, right?

n!/(n-k)!

[pale monolith]

for the first permutation you have 3! possibilities and for the second you have 3!-1
however half of these will be duplicates, so we divide by 2

[quaint flower]

I'm not sure what n is describin, that's probabaly the issue

Number of objects at your disposal?

[pale monolith]

n is the number of elements you are picking from

yes

maybe you have seen the symbol $n\choose k$

[vital dew]

Daniel

[quaint flower]

yeah I have, but I never understood it

We use:

p(n,k)

[pale monolith]

it's read "n choose k"; it's the number of ways to pick k objects from a group of n

${n\choose k}=\frac{n!}{k!(n-k)!}$

[vital dew]

Daniel

[quaint flower]

oh, we use:

n!/(n-k)!

i never seen that definition

[pale monolith]

that's something else

[quaint flower]

Isn't that the number of permutations though?

[pale monolith]

the k! ensures that picking A and then B is the same as picking B and then A

no there are n! permutations of n objects

in our case, we have $3!=6$ objects (all the permutations), and we want to pick $2$ of them. So there are

[{6\choose 2}=\frac{6!}{2!4!}=15]

ways of doing that

[vital dew]

Daniel

[quaint flower]

weird... I have never encountered that

[quaint flower]

This is what I am talking about ^^(image)

[pale monolith]

oh

[quaint flower]

yea

n!/((n-r)!)

should've been clear

oops

[pale monolith]

this is the number of orderings of $r$ objects if you have $n$ at your disposal

[vital dew]

Daniel

[quaint flower]

So for the lock, you have 10! at your disposal. If you have 10 digits to use

But, then we need to divide by something

that R

I assume it's 4?

10!((10-4)!) = 10!/(6!)

[quaint flower]

We call that permutations

[pale monolith]

for the lock you don't have 10! at your disposal

you have 4*10 things at your disposal, because each digit can be used four times at most

[quaint flower]

No I mean, for a 4 digit lock. With 10 available numbers

On each digit

[pale monolith]

so each number can only be used once?

[quaint flower]

I'm not sure.

xD

[pale monolith]

you should find out

but tbh combinations on a lock is not something you count by considering permutations

it's just multiplication principle

[quaint flower]

I know, but I am supposed to understand permutations

it's part of my course

[pale monolith]

there are better ways of understanding permutations

[quaint flower]

...I mean...

I guess? But, I don't understand.

If I have a lock of 4 digits:

A) How many times can you set different digits? Where, combination matters.

[pale monolith]

wdym write different digits

do you mean how many different possible ways of locking the lock there are?

[quaint flower]

yes

[pale monolith]

there are 10^4

[quaint flower]

Problem is,

There are two answers

And my brain hurts from reading this

In permutations, order does matter.

1110 is not the same as 0111

[pale monolith]

in the first case, you are not allowed to use the same digit multiple times

in the second you are

[quaint flower]

Are we talking about: 1110, 1110?

those are duplicates

It's just super confusing at this point

[pale monolith]

no

in the first case where there are 5040

you are not allowed to write e.g. 1120

because 1 is used twice

you are only allowe to write e.g. 1532

[quaint flower]

So, each digit needs to be unique?

problem is, my book does not talk about this shit and I am left confused which method to pick

[pale monolith]

yes

[pale monolith]

it depends on the problem statement

in the first one they explicitly say "with no digits repeated"

[quaint flower]

They do not say this

[pale monolith]

yeah they do

[quaint flower]

and yet, they answer with that lower number

how tho

[pale monolith]

read it again

"then, with no digits repeated the possible number of permutations = 10!/6! = 5040"

[quaint flower]

How many 4-digit codes are there with different numbers?

Is literally the question

[agile magnet]

wiht different numbers = unique digits

for example if you have 1231, you do not have different numbers since the thousands and ones digit are repeated

[quaint flower]

Oh, I assume it means that:

You have 10 elements, because 10 digits can be used.
You are choosing 4 elements from 10 elements

or something like that

Therefore:

(10!)/((10-4)!)

[quaint flower]

well it said so, but in a different way

[cedar lily]

v(n)=n*v(n-1)+1

can any help me solve this reccurence relation

using exponential generating functions

[half bane]

i am asked to prove this for all sets

now

what is that even mean

[blazing rose]

Is the cardinality of a group always uneven?

Since inverses come in pairs and neutral element inverse is itself?

[odd heart]

Z_n (with addition modulo n) is a group for every n

The neutral doesn't have to be the only element that's its own inverse

In {0,1} with addition modulo 2 both 0 and 1 are their own inverses

[blazing rose]

Ah so inverses aren’t unique

Right

I See

[odd heart]

They are unique, but the inverse doesn't have to be different to the original element

[blazing rose]

Yea ok

True

[odd heart]

They're unique in the sense that every element has exactly one inverse, but that might be itself

[blazing rose]

Yes

So in the case that e inverse is e

Wait no

Nvm

[odd heart]

Yeah, for the neutral element the inverse is always itself

Sometimes that's the only element with such property, sometimes not

[blazing rose]

Yea true

I get it now thanks a lot!

[brave rock]

No. An element and its inverse may not be distinct.

Nvm that message was older than I thought…

[blazing rose]

All good haha

[calm chasm]

For those who had taken discrete math, how was your experience?

[errant bear]

third easiest class i took behind macro and intro python

[brave rock]

Fun

[tribal rune]

number theory=pain for me

havent taken discrete math tho still in middle school xd

[calm chasm]

Really? At my school, I've only seen/heard negative reviews. I'm happy to hear someone had a good time

I started a bit early on my own and I feel fine.

[austere cipher]

Is "repeat the process" in this proof vaild?

[austere cipher]

Should i say
Suppose there exist some terms t such that x t is derivable from C_v and x∈var(t)
then proving any x t' such that t'=ft_1...t_n
derived from x t is also satisfying x∈var(t')?

[upper wraith]

how hard is discrete? I passed calculus 1 with an A and i'm planning to take discrete along with calculus 2

next semester

[brave rock]

Depends on the prof, and depends on you.

[upper wraith]

I’m trying to aim for an A which is 94% or above and there are 4 tests which are 80% of the grade the final is optional but if you take it it replaces your lowest score

Idk if I should take it or not I’m pretty comfortable with math like calculus but this is a new math

[vital dew]

neutch_

[calm chasm]

Can someone help explain the highlighted part and on?

[brave rock]

Let x \in R. Then there is a unique integer m such that...
So to show this definition is correct, one must actually prove that there really is a unique integer with this property.

If you keep reading, this is said very explicitly in the text.

[calm chasm]

Which definition? Also, where? I must have missed it!

[brave rock]

It's on the very same page.

The thing you highlighted even gives you the numbers.

[tribal rune]

its harder

[errant bear]

calc is way harder imo, you have so much to memorize

discrete/proofs is just logical thinking with some structure

[tribal rune]

once u do its easy 75% plug and chug if u understand what to plug

[olive ivy]

Hello I need some help figuring out how to show that the relation I was give has symetrical and then transitive property to show it is an equivalence relation (which I can see easily from the digraph and the table drawn in the picture 1).
Every other relation I was given and shown by the professor was much simpler where it was in e.g.
a^2=b^2 and then b^2=c^2 so we deduce a^2=c^2 over b^2

[foggy sentinel]

@harsh frost i didn’t pass.. I’m gonna kms😭

[cobalt bane]

excuse me, how do I solve this problem?

Given that $a_n = \sqrt{a_{n-1} + a_{n-2}}$, where $a_1 = 1$ and $a_2 = 2$. Find the explicit expression of $a_n$.

[vital dew]

QwertY

[calm chasm]

It's not clicking for me..

[agile magnet]

All they've given is a definition

You can define whatever you want

But you have to show that the thing you define does actually exist

[calm chasm]

Following so far

[agile magnet]

I can define E as the set of even primes greater than 2 and talk about their properties but you and me both know that the set E is empty

[calm chasm]

I don't think I know how to show that a integer exists.

[calm chasm]

Wait why is it empty?

[agile magnet]

Are there any even primes greater than 2...

[calm chasm]

Doesn't every set have an empty set? Or is that different

[agile magnet]

Every set does not have an empty set

∅ is not an element of the set {1, 2, 3}

For example

[agile magnet]

I would probably guess they show that in the next pages, they haven't actually shown that proof in what you sent

In fact they explicitly say "we will prove this in Theorem 6.17" so it'll be done later

[calm chasm]

I could have sworn that was true. Okay nvm.

[agile magnet]

I mean do you see why that isn't true

[calm chasm]

I see that the set consists of 3 elements. 1,2,3.

[agile magnet]

Are you confusing ∈ and ⊆?

[calm chasm]

I know the difference

[agile magnet]

It's true that ∅ ⊆ {1, 2, 3}

[calm chasm]

The empty set is a proper subset of the set you showed?

[agile magnet]

Yes

[calm chasm]

That's how to read that right

Okay nvm then.

[agile magnet]

No what I wrote is subset

Not proper subset

But it is true that the empty set is a proper subset

But I just wrote subset

[calm chasm]

Oh well this is trivial. Yeah

Baha

[agile magnet]

Anyways

Getting back to the topic at hand, do you see what the green highlighted part is saying now

They defined a thing

But they also stated that it exists and is unique

Which needs proof

Which they say they will do later

[calm chasm]

Well how about the part where it starts with "We think this ought to be true....."

I don't know why integers ought to have those properties

[agile magnet]

We haven't proven this, but do you believe that every real number is within distance 1 to an integer?

Both above and below

Like I give you a random real number x, the interval [x, x + 1) contains an integer right?

And so does (x - 1, x] right?

Just intuitively

From what you know about real numbers, I'm asking if you believe this without proof

Proving this is non-trivial at this stage

[calm chasm]

Oh yes

Yeah

We could show that easily right

Just pick any integer ?

[agile magnet]

wdym "pick any integer" I don't see how that proves what I said

For example if I say x = 1.5

I look at the interval [1.5, 2.5)

It's not "any" integer that exists in that interval

There's one in the middle, 2

It's not "any" integer, clearly 3, 4, -10, etc aren't in that interval

[calm chasm]

Okay I don't understand something.

Gonna reread these two pages and come back.

[harsh frost]

awwww sorry to hear

[tidal tinsel]

Suppose a newly-released, efficient algorithm is claimed to simulate fair dice rolls. In this case, the intention is that each outcome of a die has a 1/6 chance of occurring. Paula claims that the algorithm is faulty, and each roll will instead always produce the same number. System 2.1. To prove the die-rolling algorithm returns the same result for each roll, Paula does the following: 1. Paula tells Victor how the algorithm will roll. 2. Victor uses the efficient algorithm to simulate the rolls of the die once. 3. If Victor’s simulated roll of the dice matches what Paula predicted, he believes her claim that the algorithm produces the same result for each roll. Otherwise, he doesn’t believe Paula, since her prediction was wrong. In this case, if the algorithm returns the same result for each roll and Paula knows this result, she should always successfully predict the outcome of the simulated die roll. Consequently, Victor would always believe her. Question 2.1 (2 pts). Suppose the die-rolling algorithm correctly simulates a fair die, returning one of six random outcomes, each with a probability of 1/6 . In this case, Paula’s claim would be false. Compute the probability that she still correctly predicts the outcome of the simulated roll, which would convince Victor that the die simulation is fault

[quaint flower]

Question, so...

If I had 10 people who will be sorted into 2 teams of 5. In how many ways can you do that? Assuming that each person is unique.

Wouldn't that be solved by:

Team A: 10x9x8x7x6x5 ways
Team B: 10x9x8x7x6x5 ways

Answer: (10x9x8x7x6x5)^2 ways?

22 861 440 000 ways, basically

[grim oxide]

no

once you've picked the 5 people for the first team the second team is already defined

[quaint flower]

So...

A: 10x9x8x7x6
B: 5x4x3x2x1

[grim oxide]

no

[quaint flower]

What?

[grim oxide]

for the first team you have $10\cdot9\cdot8\cdot7\cdot6$ choices right

[vital dew]

DaBeast

[quaint flower]

I wrote that wrong

nevermind

[grim oxide]

ok

[quaint flower]

A: 10x9x8x7x6
B: 5x4x3x2x1

And that would result into 10!

[grim oxide]

but anyway it doesn't matter which order you pick the ones for the second team

because you already know which ones are gonna be in it

and the order of the team is irrelevant

so it's just $10\cdot9\cdot8\cdot7\cdot6$

[vital dew]

DaBeast

[quaint flower]

I just don't really understand the term order not mattering vs mattering

But, is it 10! ways? Because, we have team A and B

[grim oxide]

is there a difference between the team consisting of
person 1, person 2, person 3, person 4 and person 5
and the team consisting of
person 2, person 1, person 3, person 4 and person 5

[quaint flower]

Well, that's the problem. I would want to say no

[grim oxide]

yes

exactly

[grim oxide]

no

[quaint flower]

I just can't identify what order means and how to use it

[grim oxide]

you have to think about what the problem is asking

[quaint flower]

It's always an impossible task for me.

[grim oxide]

in this case the order you list the teams members doesnt change the team

[quaint flower]

Because here's the catch:

I assume that team A and B:

In A, you can have 10x9x8x7x6 ways to sort it here.

But in B, you can sort the remaining people in 5x4x3x2x1 ways

And, I would assume, that the way you pick the teams are like: This person 1 goes to team A and this person 2 goes to team B.
I'm trying to apply the multiplication principle, but, I am just super confused

[grim oxide]

the way you sort the people in the team doesn't matter

which, btw, means you dont have 10x9x8x7x6 ways to do the first team

you have $10\choose 5$

[vital dew]

DaBeast

[quaint flower]

I am massivly confused. If you have a team of 5 people. And you have 10 people.

Shouldn't this be:

1: 10 possible choices
2: 9 possible choices
3: 8 possible choices
4: 7 possible choices
5: 6 possible choices

[grim oxide]

no, because you can pick the same 5 people in different orders

which gives the same team

so you're overcounting if you just do 10x9x8x7x6

[quaint flower]

So you have a combination instead of a permutation

[grim oxide]

no you have a permutation instead of a combination

a permutation has order

[quaint flower]

Gah, I don't even know how to use this

[grim oxide]

let's say there are just 6 people

[quaint flower]

I'm trying to apply the multiplication principle, but it just does not work.

it's like smeared all over the place

[grim oxide]

then i can pick the same team of 3 people as so
ABC
ACB
BAC
BCA
CAB
CBA

[quaint flower]

I just don't understand why I don't get this though

[grim oxide]

i can't help you with that

[quaint flower]

I've been trying to understand this for 4 hours, but I can't

there's something seriously missing

I understand the addition / multiplication principle, but I cannot apply it to permutations because it's weird. I can't explain the issue.

[midnight charm]

someone help, I dont understand what a partial order or total order is btw

[quaint flower]

But, does the order matter if you sort a team?

If it doesn't then you have to use combinations, not permutations.

[grim oxide]

i mean you could construct a situation where it does

but i would say no

[quaint flower]

If every person on that team is unique yeah

[grim oxide]

unless something else is specified

[grim oxide]

no

e.g. if the order the team is listed in defines which roles the people on the team have

then the order would matter

but if it's just a group of people, then it doesn't

[quaint flower]

It's similar to a problem I solved before

Where, you have a lock right? With 4 digits.

Each digit needs to be unique.

Wouldn't that mean, the order does matter? OR does it not matter? I'm just very very confused at this point. Because what does it mean by: "Not mattering" in this case?

if you cannot have duplicate digits, then order doesn't matter?

[grim oxide]

the order does matter on a lock

the password 1234 is not the same as the password 2134

[quaint flower]

So how can I determine if something matters vs not mattering?

I need a stable framework, not abstract reasoning

[grim oxide]

think of two examples that use the same objects but in different orders

would they be the same?

in the lock case, no
in the team case, yes

[quaint flower]

So, in other words. The way you choose determines if the order matters/doesn't matter..

[grim oxide]

i don't get what you mean

[quaint flower]

I can't even explain it lmfao

in one sentence

[grim oxide]

what does "the way you choose" mean

[quaint flower]

I have no idea

I tried to say something like:

If you pick A and then B, does it matter which order you pick them?

[grim oxide]

that depends on what A and B are referring to

if i'm making a two-person team where the only defining characteristic of a team is the people on it, then no
if it's a password, then yes

[quaint flower]

I suppose, if we are sorting stuff into different classes. Classes: A,B,C

Then order would matter, yes?

because the defining charactaristic would be what class the object is belonging to

[grim oxide]

if you sort object a into class A and then object b into class B and then object c into class C, then that's the same as sorting object b into class B, then object a into class A, and then object c into class C

[quaint flower]

Uh yeah? But in some other case:

that would mean we have 3! ways to sort a,b,c into ABC respectivly

[true venture]

Hello, let's say I have the g.f. for strict integer partitions.

A(x) = Prod_{k>0} 1 - x^k

Then I add a variable y for the number of parts.

A(x,y) = Prod_{k>0} 1 - yx^k

Obviously then taking m! For some (x^n)(y^m) gives the number of compositions of that partition, but is there a better way to write the factorial step using e^y or something?

[grim oxide]

for a you have 3 choices: A,B,C
for b you have 3 choices: A,B,C
for c you have 3 choices: A,B,C

3^3

[quaint flower]

Uh, you have 6 choices tho?

3x2

If we are talking about like, sorting abc into ABC

or did I mispeak

So:

If we are sorting a group with no requirements or conditions, then order doesn't matter.

If we are sorting a group with requirements, order matters.

[grim oxide]

i'm imagining three boxes labeled A, B and C and that i have three objects labelled a,b,c which i'm placing into the boxes

[quaint flower]

Now we're going back into multiplication principle...

[grim oxide]

yes

[quaint flower]

You see this is why I am confused.

I can't apply any stable framework here

[grim oxide]

sometimes in math there isn't a formula or algorithm

you just have to think about it

[quaint flower]

I mean if we have these classes:

A
B
C

You have three objects. The way you sort them into these boxes matter? Or does it not matter?

abc
cba

[grim oxide]

define sort them into these boxes

[quaint flower]

Agh, I can't anymore.

[grim oxide]

ok so you're asking about the number of permutations of the string abc?

[quaint flower]

yes