discrete-math

[spark oyster]

No, you're getting things mixed up

We multiply the equation 22x = 29 by -3

[nimble plover]

yea, i dont understand this

[spark oyster]

But that doesn't mean we know anything more about x

[nimble plover]

yep, i got -3(22x) = -87(mod 67)

[spark oyster]

Yeah

Now on one side you have -66x; but -66 = 1 mod 67, so that's the same as x mod 67

And the right hand side is -87 mod 67 = 47 mod 67

So as a final result you get x = 47 mod 67

[nimble plover]

that's the full answer? Or just a simplification of the original question?

[spark oyster]

In the end that's the answer, yeah

you should check that every x that is equal to 47 mod 67 fulfils the original equation

and then you're done

[nimble plover]

Now on one side you have -66x; but -66 = 1 mod 67, so that's the same as x mod 67
@spark oyster im a bit confused here, how does -66 = 1 mod 67 help here?

[spark oyster]

-66 = 1, so -66x = 1x

[nimble plover]

And the right hand side is -87 mod 67 = 47 mod 67
@spark oyster we still have 1 there, wouldn't the equation here be +-1 ?

OOOH i see

because of mod you can just -66 + 67 and it's still equal

right?

so on the side with x i can do +67x any amount, and on the other side i can do +67 any amount

and it's equal?

okay i think i see it, so on the next question i had 101x=172(mod300) so i multiplied both sides by 101 (one of my numbers) and then did (101)(101x) % 300 = (172)(101) % 300 (mod 300) to get x = 272 (mod 300) which is the right answer

thanks @spark oyster

[spark oyster]

oh sorry didn't see your last questions

but i'm glad i could help!

[patent moat]

X~Y means , Food Y is included less than food X in the everyday day menu
letters represent foods

A<B
C <D
E <F
F<G
E<H
I<G
G <K
D< L
L<G
L<B
M<C
A<L

Is this a partial provision relationship? Strict partial provision; Total layout;

im totally lost on this one , anyone has any idea?

[whole hawk]

how to compute echelon form over ring of integers mod non prime??

[weary tiger]

ok so this is a bit of a bruh moment

but i don't seem to be able to prove this

Let S be any set. Prove that the law of composition defined by ab = a is associative.

idk i just tried to prove a(bc)=(ab)c but a(bc)=a and (ab)c=ab

[opal crescent]

well what's ab

you can simplify again

[weary tiger]

,,,,

lmao ty

[pale wing]

If a graph A has a subgraph homeomorphic to graph B and graph B has a subgraph homeomorphic to graph A

Does that necessarily imply graph A= graph B?

[fervent cove]

knowing the inicial condition, how do i prove the following?

using induction atm, but not being able to prove P(n+1)

[pale epoch]

why not?

[prisma arch]

needs me a little help with this following problem for b and c. My top row for c matches the top row for the inverse of b but i cant get it to match up the bottom row.

[pale epoch]

what are your results

[pale epoch]

you did c) wrong

[prisma arch]

may gave found my mistake on that thank you

have*

[pale epoch]

also to show that two matrices are inverse of each other you could also multiply them together and show the result is the identity matrix

[whole hawk]

ax = b mod m

how to solve something like this

where a, b and x are vectors

[pliant pendant]

~~Children in kindergarten arrange towers using colored blocks. They have two types of blocks: blocks of length 1 in three colors (white, red, black) and blocks of length 2 in four colors (yellow, green, blue, brown). Determine, by arranging a recursive relationship, number of towers of height n.~~
Hello, I have a question about this relatively easy problem. Is it possible to arrange recursion with regards to colors? Without colors its easy Fibonacci sequence but I don't know how to count colors in this.
If something is unclear in my explanation let me know, English isn't my first language.
I solved it by myself. I needed to multiply a_n-1 (block of length 1) and a_n-2(block of length 2) by number of colors (3 and 4 respectively)

[whole shard]

What do you guys suggest I should review ( pre-uni) to get better with discrete math

I think im abt to fail again

[vale sedge]

Anyone familiar with Shift Registers for Shift Matrix and Generating Matrix?

[lilac pivot]

for even numbers it's obviously true

for odd numbers, $\ceil{\frac{n}{2}} = \frac{n+1}{2}}$, $\floor{\frac{n}{2}} = \frac{n-1}{2}}$

[vital dew]

PorosInMyAshe:
Compile Error! Click the reaction for details. (You may edit your message)

[lilac pivot]

$\frac{n!}{\left(\frac{n+1}{2}\right)!\left(n-\frac{n+1}{2}\right)!} = \frac{n!}{\left(\frac{2n-n+1}{2}\right)!\left(\frac{2n-n-1}{2}\right)!} = \frac{n!}{\left(n-\frac{n-1}{2}\right)!\left(\frac{n-1}{2}\right)!}$

[vital dew]

PorosInMyAshe:

[lilac pivot]

and there you go for odd numbers

just some algebra

[weary tiger]

how do i begin to find the probability that five rolled values from a 10-sided die form a weakly increasing sequence?

[lyric pumice]

@weary tiger Observe how many ways a particular distribution of values can appear.

[north jewel]

How many ways are there to arrange the letters AABBCDEF (length 8)?
I've thought of two ways

• 8C2 * 8C2 * 4! (choose location for A's, choose location for B's, then choose locations for CDEF)
• 8! / (2! * 2!) (number of permutations, then divide by the overlap?)

But they're different; which one did I mess up on?

[lyric pumice]

@north jewel The first result overcounts by giving too many possible locations for Bs.

[north jewel]

oh duh

should be 8C2 * 6C2 * 4!

I guess I'm just blind... they match up now

thanks haha 😅

[dapper rose]

This was a claim in my graph theory class

but I came up with a counterexample?

k(G) is number of connected components of G

graph means simple graph

[lyric pumice]

@dapper rose How many connected components of G are there?

[dapper rose]

in my counterexample? Isn't it 3?

<@&286206848099549185>

[lyric pumice]

S is not a vertex cut.

[dapper rose]

why?

looking at this my S looks to be a (y,z)-vertex cut

[weary tiger]

too late already done

i did some stars and bars magic

[lyric pumice]

@dapper rose The removal of S decreases the number of connected components present.

[dapper rose]

yes, which is a counter example to the claim that removal of an arbitrary S increases the number of components?

[lyric pumice]

It is not a counterexample.

It is not the case that k(G-S)>k(G). So, the requirement for {x} to be a vertex cut is not met by the definition.

[dapper rose]

you aren't making any sense, the definition of a (y,z)-vertex cut is given in definition 5. {x} satisfies:

1. {x} is a subset of V(G) - {y, z}
2. y and z are in difference components of G - {x}

I'm not saying {x} is a (x,y)-vertex cut

I'm saying {x} is a (y,z) vertex cut

Millenial what you have been saying makes no sense to me. My question still stands for any other ppl reading this

[lyric pumice]

The definition is confusing because it is incomplete.

x and y should be assumed to be connected.

Of course, strictly, it may be plausible that {x} is a (y,z)-vertex cut and yet {x} is not a vertex cut of G, but I am doubtful of this.

[dapper rose]

here's the definition our professor gave for a vertex cut, which I've been using in this conversation

[pallid lintel]

this seems like a strange proof of the handshaking lemma to me. it essentially relies on counting the rows and collums on the incidence matrix? (modified so loops are 2 instead of 1). hows that any different from counting vertices/edges on a graph by hand?

[lyric pumice]

@dapper rose What kind of professor do you have?

There is no finite graph in which every vertex is a vertex cut.

[dapper rose]

a cut vertex and vertex cuts are different things

you are looking at https://math.stackexchange.com/questions/2461202 where they are talking about cut vertices? @lyric pumice

this is my class's definition of cut vertex

[lyric pumice]

https://math.stackexchange.com/questions/3581839/do-we-only-speak-of-a-cut-vertex-when-talking-about-a-connected-graph?rq=1

[lyric pumice]

https://mathworld.wolfram.com/ArticulationVertex.html

[pallid lintel]

we didn't use vertex cut definition in our class because we can use a slightly more complicated definition that allows us to say more things. (called a seperation). a seperation is when AUB=V in G(V,E) and no edge from A-B and B-A is connected. A intersect B is the boundary, its size is the order of the seperation

[weary tiger]

how would one show this concisely

[sour arrow]

I don't see anything to show

[weary tiger]

oops

i think i got it anyways

its to show general associativity

i used induction

[stray reef]

it's not that they "could" both have a cardinality of 3

they DO both have a cardinality of 3

yes

[wary lichen]

Hi i need help with fine state acceptor

[wary lichen]

turning my codes to deterministic keep going on and on what should i do

[lime jolt]

Is this correct for finding the gcd?

29 = 3(8) + 5
8 = 1(5) + 3
5 = 1(3) + 2
3 = 1(2) + 1
2 = 1(1) + 1 <-- gcd

[weary tiger]

nah

the last line should be 2 = 2(1)

@lime jolt

for instance

if you had 14,4

14=3(4)+2

4=2(2) <- gcd

[lime jolt]

oh damn, thank you @weary tiger

[lime jolt]

Can someone help me with Eulers Theorem with 7^55320486

[faint narwhal]

[lime jolt]

How do i calculate the last two digits of that huge number using eulers theorem

[faint narwhal]

can you think of a way to transfer "last two digits" into something involving mod?

[lime jolt]

is it like this

7^55320486 = 7^5532048 + 6 = 7^10(5532048) * 7^6

= (7^10)^5532048 * 117649 = 1^5532048 * 117649

?

honestly im not too sure

[faint narwhal]

uh

[lime jolt]

following some method my teacher did

[faint narwhal]

but why is 7^10 = 1? in the last step?

Thats not true normally

[lime jolt]

I have no clue lol

In all the examples, he gave us a mod

but in this problem he did not provide it to us

so I am not sure how to get it

[faint narwhal]

Well again, like I said

can you think of a way to transfer "last two digits" into something involving mod?

[sterile sandal]

would n_b(x) here be the number of times 'b' is in the string?

[static basalt]

and there you go for odd numbers
@lilac pivot THanks just saw this now

[queen patio]

I can't really think of a better place to ask but

can any language be defined as the limit of a sequence of regular languages?

hm, actually the answer is yes because any language can be defined as a limit of finite languages and all finite languages are regular

nevermind

[hidden geyser]

how would one prove this

[last sigil]

Consider the numbers mod 4 @hidden geyser

You can try to generalize the result afterwards

[hidden geyser]

Are you telling me to use the numbers -3, -2, -1, 0, 1, 2, 3 to formulate the proof or that there's some insight to those numbers?

[last sigil]

Consider only positive residues mod 4

0, 1, 2, 3 - Notice we have 4 residues and 5 numbers in our subset. What does the pigeonhole principle tell you now?

we ignore -3 mod 4, since this is equal to 1 mod 4; same reasoning for other negative residues

[hidden geyser]

The difference doesn't hold in the case of a 4 element set is what I'm getting?

Sorry if I'm not getting something obvious I'm a programmer not a mathematician 😄

[last sigil]

This does hold

Whats this for btw

[hidden geyser]

It's a past exam paper from 2017 for our maths module.

[last sigil]

Also, can you restate the pigeonhole principle? Just wanna make sure you understand it

[hidden geyser]

If we have an excess of items from containers at least one container must have more than 1 item is my understanding of it.

[last sigil]

Yeah, kinda awkward how you stated it though

So basically we have "boxes" and "objects"

If we have more objects than boxes, then at least one of the boxes will have more than 1 object

Now when using pigeonhole, the tricky part is finding out what our boxes & objects are

0, 1, 2, 3 - Notice we have 4 residues and 5 numbers in our subset. What does the pigeonhole principle tell you now?
Going back to this, what do you think the boxes & objects might be?

@hidden geyser

[hidden geyser]

no clue

[last sigil]

Alright, what do you know about a-b mod 4

[hidden geyser]

will be 0, 1, 2 or 3

[last sigil]

No?

[hidden geyser]

oh will be a - (0, 1, 2, 3)?

[last sigil]

Alright, another direction

We need to determine our boxes and objects

[sleek swallow]

yea in a direction straight to your mom

prank

[last sigil]

Hint: Our boxes are the residues mod 4

So what do you think the "objects" are?

[hidden geyser]

define residue

[last sigil]

Note we have 4 boxes

residues mod 4 are (0, 1, 2, 3)

Every integer mod 4 must be 0, 1, 2, or 3 mod 4

[hidden geyser]

I'm asking what residue means in this context not what they are

[last sigil]

residues are the possible remainders after dividing by 4

[hidden geyser]

Why are they the "boxes" in this situation?

They don't seem like they could contain anything

[last sigil]

Our "items" will be the elements of your set

You have 5 items now right?

And 4 boxes

So one of the boxes will have 2 elements of the set that are the same residue mod 4

[hidden geyser]

so are the boxes those numbers and the objects are how many times each appear when we mod the numbers in the set by 4?

[last sigil]

yeah

well, kinda, your explanation isn't quite right

[hidden geyser]

then what would be correct ?

[last sigil]

So we got our 5 element subset {a, b, c, d, e}

And 4 boxes, which are 0 mod 4, 1 mod 4, 2 mod 4, and 3 mod 4

Since every integer is 0, 1, 2, or 3 mod 4

And we have 5 elements

Pigeonhole tells us that one of these boxes contain 2 elements of the set

You following?

[hidden geyser]

ye that's what I said

[last sigil]

alright

Can you finish the proof from here?

[hidden geyser]

Could I ask why do you ignore negative numbers? And how this would work in the case where we have 5 numbers that are say something like 8, 12, 16, 20, 24?

those would just go into 0 5 times

[last sigil]

Sure that still works

Take 12 - 8

That is a multiple of 4

Infact, notice any combination of the difference of those is a multiple of 4

[hidden geyser]

oh yeah sorry got a bit tunnel visioned there

[last sigil]

Reason we can ignore negative numbers mod 4, is that -3 mod 4 is equivalent to 1 mod 4

[hidden geyser]

that's not how I know mod to work

[last sigil]

How do you think mod works

[hidden geyser]

-3 % 4 = -3

[last sigil]

a mod b= a + b mod b

Also

-3 mod 4 = -3 is saying -3 is 3 less than a multiple of 4 (0 - 3)

But

-3 is also one more than a multiple of 4 (-4 + 1)

[hidden geyser]

that's how modulus works in any programing language

[last sigil]

Sorry, then

This doesn't change the fact they are equivalent

[hidden geyser]

I take it for -5 % 4 => -1 % 4 => 3 % 4 = 3?

[last sigil]

yeah

-5 is 3 more than a multiple of 4 (-8 is the multiple; -8 + 3 = -5)

Is that clear now?

[hidden geyser]

yeah maths mod =\= programming mod

lol

[last sigil]

Sure

Alright, back to the main point at hand

Note that pigeonhole tells you one of the boxes contain at least 2 objects

(this covers the case where all 5 are 0 mod 4)

[hidden geyser]

a and b are those 2 objects is what I'm leaning towards

or

any pair that are in the same box

[last sigil]

Yes

Can you prove why the difference of any pair in the same box is a multiple of 4?

[hidden geyser]

(4a' + n) - (4b' + n) 
n = common residue
a' = (a - n) / 4
b' = (b - n) / 4```

and then simplify?

[last sigil]

yep

So the proof is essentially complete

See how pigeonhole helped us here

[hidden geyser]

Thank you so much for being so patient 😄

[last sigil]

It established that at least 2 of the elements share the same residue mod 4

Lets call them a and b

We then had to simply show a -b mod 4 = 0, which implies a -b is a multiple of 4

@hidden geyser Using a similar argument, one can show that in any n+1 -element subset of the integers, There exists a and b, such that a-b is a multiple of n

I encourage you to try it for yourself as practice

[hidden geyser]

seems simple enough

just add another variable instead of 4

2 actually

[last sigil]

There are a lot of cool pigeonhole problems out there; recognizing the boxes and objects are usually the toughest part

Anyways, np; good for you for hanging in there!

[hidden geyser]

Until I'm stuck again

[solemn cairn]

http://puu.sh/FFQ3f/4d130f4734.jpg

no idea how to start, any help?

[pale epoch]

look up what Big-$\Theta$ runtime means

[vital dew]

Lochverstärker:

[solemn cairn]

oh nvm i got it thanks lol

overthought that

[mighty garden]

<@&286206848099549185> need help in my math homework, it's sets and functions. anyone can help me?

[lime turret]

@solemn cairn Take the dominant term to find Big O when you have a complex term thrown together, usually works.

Looks like you might want to do some distribution first though, hint hint

post it here? or DM I can try @mighty garden

[mighty garden]

I already got it

Thanks though

[quartz gull]

<@&286206848099549185>

[eternal patrol]

well

what do you think?

also

!15m

[weary tiger]

use Bayes' theorem or just work it out directly

[weary tiger]

@quartz gull

[spark oyster]

that ain't a good name fam

[weary tiger]

<@&268886789983436800> 👀

[toxic zodiac]

Would rationals and irrationals be a proper subset of all real numbers?
A proper subset means that one set MUST have some elements of the other, and is not allowed to be equal. So, if that's the case, then all real numbers CAN be represented by rationals and irrationals. Therefore, rationals and irrationals would NOT be a proper subset, as they would be a (regular) subset.
Would this assessment be correct?

Written in equation form,

thereby making this equation incorrect

[pale epoch]

irrational numbers are by definition real numbers that are not rational

so the union of all rational and all irrational numbers is by definition all real numbers

notice that some some authors use $\subset$ instead of $\subseteq$

[vital dew]

Lochverstärker:

[toxic zodiac]

ah yeah--so this is one of those things where the definition of subset and subseteq matter based on asker 🤦

but I understand it! thank you!

[foggy scarab]

can someone break this down for me? I understand recursion but I'm having a hard time understanding what all the notations mean

[obtuse minnow]

Tell me what ya know about induction.

[foggy scarab]

i have to prove n+1 and n-1 to be true for the induction to be valid

[obtuse minnow]

So tell me what you know about the induction step?

(either in general or specific to this problem or both)

[foggy scarab]

usually i take the equations given in the problem and manipulate it to where it is equivalent to the question being asked

but im mainly having trouble understanding what all the math symbols mean

[obtuse minnow]

ah okay. (a, b) is a two-coordinate object.

The first line states (1, 5) and (3, 7) are both two-coordinate objects in S.

[foggy scarab]

so theyre just coordinates? not like a range such as 1-5 and 3-7?

[obtuse minnow]

Then next line states there's other guaranteed objects.

I am pretty sure they are coordinates in this case

[foggy scarab]

so the recursive step is saying if there is one coord, follow the equation. if there are two coords, plug them together and make them into one coord?

[obtuse minnow]

it is better than that, for ANY object you have, then we'll also include the modified object.

for ANY 2 objects, we will include the formula for a new object and that object will be in S.

It's a bit like this: I have two numbers (not coordinates just two separate numbers) 5 and 10 in T.

I tell you whenever you have a number in T I will promise you that half of that number is in T.

Can you show me there are (some property) of elements in T. I could go for:

1. there are infintely many elements between 0 and 1.
2. infinitely many terminating decimals.
3. no repeating decimals. 🙂

I'm trying to provide extra examples

But yes, the second line with (c, d) says exactly that.

If two coordinates are in S combining them in this given way results in an object in S

[foggy scarab]

ok so i assume i will need to use both recursive steps in my proof?

[obtuse minnow]

Yes but one at a time.

Like you want to prove every possible new thing as defined by the top must also be in the set.

then you have to prove every possible thing as defined by the bottom will also be in the set.

[foggy scarab]

so what im thinking right now is to start proving two coord, then one. since the example im trying to prove only gave me one coord, ill have to convert it to two coord? does that sound right?

[obtuse minnow]

So for the second part it says IF these two things are in S then so is (blah). Can you set up the condition (if part)?

Basically you say if (a, b) is in S and (c, d) is in S, then a + 4 = b and c + 4 = d.

Then show that (2a + d, 2b + c) follows the pattern for S.

[foggy scarab]

ok i believe i proved that above, and reduced (2a+d, 2b+c) to y = x+4. does that mean i proved it for two coord and do i do the same for one coord using (3a + 1, 3b − 7)?

[foggy scarab]

so i was able to get both (2a+d, 2b+c) and (3a+1,3b-7) to equal y= x+4. does that mean i have successfully proven the recursive definition?

[proven girder]

Could someone correct me if I'm wrong with this? (Please):
I don't think this is possible, since there's only 5 people there can only be 2 pairs. Like the pigeonhole principle, one of the handshakes will have to repeat which isn't allowed. Therefore, it isn't possible.

[pale epoch]

you are correct but i dont get your argument

what are the pigeons, what the holes?

[proven girder]

There are more values than there are variables, so one of the values (handshake) would have to repeat.
I'm assuming that in order for this to work there would have to be 2*3 people in order for there to be 3 unique handshakes.

Thank you for confirming it by the way, I've fallen for some tricky questions would rather not try to lose a few points

[pale epoch]

ok, you have the right idea i think

but i dont see how to turn this into a pigeonhole argument

like, the number of handshakes is the pigeons and the boxes are what exactly?

(there is an easier argument if you covered the handshaking lemma)

[proven girder]

I was thinking that because there's a repeating value that it could be tied to that principle, I'll just keep it out of the argument since I myself don't even know that. But for some reason it did help me understand it a bit easier.

[weary tiger]

@proven girder do you understand a solution now?

Because if you want a relatively simple solution

[proven girder]

I do understand it, but I'd like to hear the simple solution!

[weary tiger]

I'm not sure how simple it'll be but

WLOG let A shake B,C,Ds hands

Then E must shake B,C,Ds hands also because if not A has shaken more than 3 hands

So we have 3,2,2,2,3 being the number of hands shaken of A,B,C,D,E respectively

[little matrix]

can u gimme the answer someone plz

[weary tiger]

Then WLOG B shakes Cs hand so you end up with 3,3,3,2,3 and clearly it can't be done

@little matrix angles on a line add up to 180 and angles in a triangle add up to 180

you can do it from there

[little matrix]

that adds up to 181

all the angles line up to that

[pallid lintel]

Is this a good definition?: A walk in a graph is an alternating sequence of vertices and edges v1e1v2e2···en−1vn such that each ei is incident with both vi and vi+1

i feel like it should be specified that vi can equal vi+1 for loops

[obtuse lance]

your definition has some redundancy I guess, unless you're specifically allowing for multiple edges between vertices

[pallid lintel]

the course lecturer has us doing non simple graphs yes (meaning loops and parrell edges)

[mellow girder]

My Assignment

I missed my classes, what topic do i need to search up and learn to be able to complete this assignment?

[ancient otter]

Hi, can someone help me to get this?

Z_n is the same as Z/nZ the ring of residue classes mod n
Z*_n is the elements in Z_n where gcd(x,n)=1

what is nZ

[pale epoch]

the set {n*z | z in Z}

so all integer multiples of n

[weary tiger]

Does anyone know how to do mathematical induction like this example

been stuck on this for weeks

i cant seem to grasp it

at least how to start would be greatly appreciated

[stray reef]

start by writing out your inductive hypothesis and your goal

[robust fog]

can anyone help me with this integral?
been on some olympiad from 2008 and i've no idea how to wrap my head around it

[weary tiger]

start by writing out your inductive hypothesis and your goal
@stray reef what is that, never taken it

[stray reef]

you've never done mathematical induction before?

[near prawn]

@robust fog divide top n bottom by x^2

then use u=1/x-x as a sub

also wrong channel

[robust fog]

sry for that, but i still cant really work it out

[main heart]

Hi sorry, but what is the difference between an additive group and multiplicative group?

and can't an additive group be a multiplicative group as well, since the adding a value to itself is just multiplying?

I am keeping the discrete log problem and Elliptic curves in mind, when I ask this

[stray reef]

additive vs multiplicative out of context just means that the group's operation is called addition or multiplication respectively

[obtuse lance]

often times people will use + to specifically mean a commutative group, but it's not a set in stone convention

[main heart]

Aren't both group commutative? Or are things like matrix multiplication also considered a multiplicative group

[pale epoch]

the set of all invertible matrices with matrix multiplication is a non-commutative group, yes

[main heart]

oh ok... yeah, makes sense

thank you to all of you!

[pale epoch]

also i wouldn't really use the term "additive group"

and "multiplicative group" refers more to the multiplicative group of a field specifically

ultimately the symbol you use to denote the group operation is arbitrary

[main heart]

I see... just have to be aware on how I use them, i guess

[lapis geyser]

Hello, I'm trying to figure something out here. I need to justify if a function is 1-1 and/or onto. I understand 1-1 but not so much onto. In this problem I have where f: Z x Z -> Z defined by f(x,y) = x + y . I know it is not 1-1 because f(1,0) = f(0,1) but (1,0) does not equal (0,1). For onto, I am currently going with it is onto but because f(0,y) = y. I'm not sure I get how to explain/justify how something is onto or not.

[pale epoch]

it's onto if every element in the codomain has a preimage

like in this case: for every z in Z, you have to find a (x,y) in Z x Z, such that f(x,y) = z

which you did

[lapis geyser]

So technically the explanation that it is onto because f(0,y)=y is correct?

[pale epoch]

not just technically

it's correct

[lapis geyser]

Okay thank you. Now what happens if I explained it f(x,0) = x would that be incorrect?

[pale epoch]

no

you just have to find any preimage

in this case every element in Z has at least 2 preimages

but finding one is enough to show the function is onto

[lapis geyser]

Oh okay cool.

Thanks

[lyric pumice]

@mellow girder Look at combinatorics and sets.

[lapis geyser]

When doing inverses of a floor or ceiling function can you end up with something like [sqrt(-10), sqrt(-5)) or is this something considered imaginary and is not included in the final answer of an inverse

To explain more f^-1(s) = [sqrt(-10), sqrt(-5)) U [0, sqrt(5)) . Could the final result be f^-1(s) = [0,sqrt(5))?

[near prawn]

@robust fog did u figure it out

[wintry rock]

Does this look like a correct DFA to you? Reg expression is ab(ab) + c

I kinda used my wits lol

Whoops

That's the reg expression

[novel quartz]

Sorry if this is against the rules but is there anyone I can pay as a tutor to guide my thinking for an assignment I have due in 10 hours

[last sigil]

Yes; Payment is against the rules

[novel quartz]

understandable

[main heart]

quick question... but what does it mean when a statement says (z/5z)^2 as a finite field

does it mean that exclude all integers which are multiples for 5 and square the remaining?

Sorry if this is the wrong channel

[stray reef]

no that's not what that means

[main heart]

then how do I interpret the symbols?

Ann to my rescue once again

[stray reef]

Z/5Z is the field of integers modulo 5

[main heart]

ok... so / implies modulo?

[stray reef]

it's the notation for quotient groups and rings

[main heart]

oh alright

[weary tiger]

Is anyone here familiar with extended euclids

sm+tm

[olive hamlet]

i think they meant $A\in P(B) \iff A\subset B$

[vital dew]

JohnDoeSmith:

[stray reef]

rho is their relation symbol

the partial order in your poset

[fallow bronze]

I have a question on graphs: Let G be a graph with at least 4 vertices. Suppose that each vertex in G has degree of at least 2. Prove that there is a simple path in G of length 4. For some reason I have no idea how to prove this. Any help?

[obtuse lance]

since all vertices have degree 2, every vertex you enter connects to another vertex, so try constructing a simple graph with only paths of length 3 or less

[pale epoch]

is & set union

then ye

no idea where the 40 comes from

but ye

this is called the inclusion-exclusion principle

[fallow bronze]

Does anyone know how many connected components this graph would have?

[stray reef]

one

[fallow bronze]

Why is that?

[stray reef]

this is a connected graph

for any two vertices in the graph there is a path from one to the other

[obtuse lance]

start from one vertex and start coloring edges the same color if they're connected until you run out, that's one simple way to do it

[fallow bronze]

Oh I see. Thank you. I will try out this coloring method too

[atomic steeple]

What I have so far is "the equivalence class for each integer includes every real number with a floor value equal to that integer"

but is there a fancy notation way to write this

[obtuse lance]

write it in terms of an interval

[stray reef]

the equivalence classes of this relation are intervals of the form $[n, n+1)$ where $n \in \bZ$

[vital dew]

Ann:

[fallow bronze]

I have a problem similar to this on my homework but I would rather know how to do this so can anyone run me through the process of how to solve this? For example: Theres a total of 20 distinct cards. I have 14 distinct cards and my friend has 11 distinct cards. How many do we have in common, at least?

[stray reef]

you're missing some underscores

you're still missing some underscores!

[vital dew]

work:

[stray reef]

anyway, if $x_i$ denote the indegrees and $y_i$ the outdegrees, then $\sum_{i=1}^n x_i = \sum_{i=1}^n y_i$

[vital dew]

Ann:

[stray reef]

yes

exactly

you're kinda overthinking it

$ \sum_{i=1}^n (x_i^2 - y_i^2) = (n-1) \sum_{i=1}^n (x_i - y_i) $

[vital dew]

Ann:

[stray reef]

you've proven this

and you've proven $\sum_{i=1}^n (x_i - y_i) = 0$

[vital dew]

Ann:

[stray reef]

not everything needs induction lmao

$\sum_{i=1}^n x_i = \sum_{i=1}^n y_i = |E|$

[vital dew]

Ann:

[stray reef]

this is because each edge contributes 1 to the indegree count via its head and 1 to the outdegree count via its tail

this is true in any directed graph

yeah so that proves $\sum_{i=1}^n x_i = \sum_{i=1}^n y_i$ and hence by a simple algebraic manipulation $\sum_{i=1}^n (x_i-y_i)=0$

[vital dew]

Ann:

[white monolith]

does anyone here know how to do relations and induction proofs

im a little confused

[near prawn]

just ask

[minor crest]

Hey could someone else help me pls

[pallid lintel]

whats this question mean by 'G is an odd cycle'? Does that mean all the vertices in G are connected in G to form an odd cycle?

[ripe cave]

Yeah, to be presice it says that G is isomorphic to C_n for some odd n

In other words, the vertices of G in some order form an odd cycle and there is no edges in G other than edges of that cycle

[pallid lintel]

right, thats what i thought, thanks

seems like i can prove it by showing that any 2 odd cycles joined togethor make an even cycle if i try to colour them. odd+odd length paths=even length.

[pallid lintel]

maybe represent it as a balls into urns problem. set circuit with 6 vertice as 1ball {a}, 7 vertices as 2ball, {a,b}, ect.

you done permutations and combinations before?

i think your on the right track.

circuits are paths that end on same vertex right? you can't use other vertex more than once so i would have thought a length 3 circuit is 10*9 * 8

length 6 would be 10x9x8x7x6x5

ah ok.

[ripe cave]

Are you sure? The common definition of circuits specifies that the edges can't repeat

[pallid lintel]

all my 3 graph theory courses ive done havnt even mentioned circuits. definitions seem to change from teacher to teacher

[ripe cave]

Lol yeah, you should check the presice definition that was given to you

Also sometimes the circuit 1-2-3-1 is identified with 2-3-1-2 and 3-1-2-3

[pallid lintel]

mindfucking me haha, my definition of path must be different than yours

yeah dont listen to me. i've been using different definitions to you

[sterile scarab]

what defines discrete math?

what is recurrence relation?

ah

[woeful galleon]

A recurrence relation is an equation that expresses each element of a sequence as a function of the preceding ones i know what it is but this is straight from wiki (i do coding so hard to explain)

[sterile scarab]

the sad thing is im in HS and im doing shit covered in Calc III

not for school, but because I decided to torture myself

well

[woeful galleon]

that is ^

[sterile scarab]

im implementing a research paper that discusses topics part of calc 3

such as vector gradients, tangent planes, normals, etc

mainly applied into computer science specifically

and mean squared error

I guess thats more stat then anything tho

lol

theres the main equation

I understand how to find a vector gradient

but how do I find the tangent plane?

dont I need normals for a tangent plane?

and how would I find the pseudo inverse of that function

the paper suggests if there is no solution to just use the pseudo inverse?

oh

the pseudo-inverse of the matrix

im really dumb

sorry guys

https://www.youtube.com/watch?v=nS3qwFOPieg

yay I found something that might help me

[pallid lintel]

if you take a circuit (1,2,3,1) is the same as (1,3,2,1) backwards? in any case maybe your lecturer got the question wrong for not being specific enough with definitions! They are definitely different on weighted and directed graphs but not sure about simple graphs. will have to ask your teacher i think

[fading abyss]

Hi , this isn't really a help post but im taking Discrete next semester and I was wondering if anyone had any suggestions on books/lectures/videos/practice exams(esp with solutions)

[pallid lintel]

mit ocw is usually good. books are pretty indepth...a lot more problems in them than the course usually covers.

[blissful ibex]

How can i prove the inequality of 2 combinations?

Second task

[stray reef]

[moved to #help-2]

[bronze sky]

i have no idea what this question wants me to do in my head

could anyone help me with this?

[tulip vale]

to clarify does $\overline{A}$ mean the complement of $A$ in $U?$

[vital dew]

EGA 1:

[bronze sky]

yes

i cant stop thinking that A has to be Ø, or something

cause in my head it doesnt make sense

[tulip vale]

okay what it says is that $A \neq U,$ and $B \neq = \emptyset.$

[stray reef]

no?

[tulip vale]

A centernot = U

[stray reef]

since when did B have to be empty just because $\overline{A} \cup B = U$?

[vital dew]

Ann:

[tulip vale]

A and B centernot those things lol

[stray reef]

what's "centernot"

[bronze sky]

ye

[stray reef]

did you mean the not-equal symbol

[tulip vale]

not equals

[stray reef]

that's \neq

$A \neq U$ and $B \neq \varnothing$

[vital dew]

Ann:

[tulip vale]

yes

that's what I was trying to type lol

[stray reef]

anyway $B$ need not be nonempty

[vital dew]

Ann:

EGA 1:

[bronze sky]

what

[stray reef]

indeed if A and B are both empty then the equality $\overline{A} \cup B = U$ is true!

[vital dew]

Ann:

[stray reef]

what you can say is that $A \subseteq B$, which imo is a much stronger statement

[vital dew]

Ann:

[bronze sky]

ok but how did you get to that conclusion

[stray reef]

$\overline{A} \cup B = U$, intersect both sides with $A$ to get $(A \cap \overline{A}) \cup (A \cap B) = U \cap A = A$

[vital dew]

Ann:

[stray reef]

$A \cap \overline{A} = \varnothing$ so the left-hand side is just $A \cap B$

hence $A \cap B = A$, which is equivalent to $A \subseteq B$

[vital dew]

Ann:

Ann:

[tulip vale]

oh shit I missed for all x

I though it was there exists x

so I just wanted the thing to be nonempty

[bronze sky]

so A = Ø?

[stray reef]

no

where did i say A was empty?

[bronze sky]

A intersects not A = Ø

what does that mean

[stray reef]

it means exactly what's written

the intersection of A with its own complement is empty

[bronze sky]

yes

[stray reef]

which should be obvious if you know what a complement is and what an intersection is

[tulip vale]

All you need is A is contained in B

[bronze sky]

yeah thats whats cinfusing me, why do you use it

[tulip vale]

because Everything not in B will then be in the complement of A

[stray reef]

ega 1 you're kinda overcomplicating it and at the same time repeating what i said already

i took the condition $\overline{A} \cup B = U$, which is given, and from it derived that $A \cap B = A$

[vital dew]

Ann:

[bronze sky]

fuck

how did you derive that

am i stupid

got it

thank you @stray reef

I was stuck on this one, how do you see that connection so fast?

[balmy nacelle]

hey you guys, I'm stuck on a homework question right now

don't they all have the ability to be false?

[stray reef]

@bronze sky ¯_(ツ)_/¯

@balmy nacelle no, a is true

[bronze sky]

3a is true

why are there to 3a's?

and b¨s

b's

[balmy nacelle]

my professor makes bad homework assignments lmao

which a are you guys talking about, the first one?

[stray reef]

lowercase a

some letter will be used more than twice

there's 10 questions

if each letter is used twice or less, that only covers 8 questions at most, which is less than 10

[bronze sky]

thats the only one thats true

[tulip vale]

the second b is also true; with only ten questions, there can't be at least 3 answers with all 4 choices

[bronze sky]

and what is the test is just A's

like where all answers are the A's

[tulip vale]

that's fine, then b won't be used more than 3 times

it doesn't say every letter will be used at most 3 times

[bronze sky]

ohhh

i didnt get the question

[tulip vale]

it just says "some letter" will be used at most 3 times

[bronze sky]

mb

[tulip vale]

this is a terrible question lmao

it has two part a's and two part b's

[bronze sky]

lol ye

[tulip vale]

So the only ones that have to happen are the first a, and the second b, @balmy nacelle

[bronze sky]

wait i seriously cant wrap my head around 3 2nd b

doesnt it say that any letter will be used 3 times max?

[tulip vale]

no is says some letter will be used at most 3 times

which means: there exists a letter which is used at most three times

it doesn't say: each letter will be used at most 3 times

[bronze sky]

ohhhhhhhhhhhhh

okok

[spark oyster]

the second b is not necessarily true, because it's a multiple choice test

[bronze sky]

yeah i get it now

welp

doesnt that mean all letters can be correct in a question

[spark oyster]

yeah

all questions can have all letters as correct for example

[bronze sky]

i overlooked that

[tulip vale]

Lol I assumed that only one answer is correct for each question

[spark oyster]

it's sneaky

[bronze sky]

damn

yesh

so 1st3a is false

1st3b is false

all are false i guess

[tulip vale]

the first a is still true; because you have to write down at least 10 letters

and you're choosing from only four of them

so you have to repeat

[bronze sky]

i think they mean the correct letter

like when you are done with the test you look at wich letter was used, assuming you 100% the test

[tulip vale]

Oh so you're saying you think that the combination AB as the correct answer does not count as "using an A"

[bronze sky]

its using a in that case

i mean the test can go:
A : Correct

fuck

wait

q1
A : Correct
B : false
C: False
D : false

q2
A : Correct
B : false
C: False
D : false

and repeating

[tulip vale]

Sure

[bronze sky]

that means B C D was never used

[tulip vale]

that's fine; you used a more than twice

[balmy nacelle]

are you guys good at proofs?

[tulip vale]

it says "some letter" not "each letter" lol

[bronze sky]

uogtvrae

[tulip vale]

Depends on what "good" means lol

[bronze sky]

jesus i swear to god i have dyslexia

[balmy nacelle]

I literally hate proofs lol

[tulip vale]

I'm not top 50 putnam good

[balmy nacelle]

so for the base case

[tulip vale]

The base case just plug in n = 0

0 < 1 checks out

Also check $n = 1,$ $2^1 = 2(1) = 2,$ so it checks out. Suppose that $2n \leq 2^n,$ $n \geq 1$ we w.t.s. $2(n+1) \leq 2^{n+1}$ But then, since $2(n+1) = 2n + 2,$ and $2^{n+1} = 2(2^n) = 2^n + 2^n,$ we apply inductive hypothesis to say $2n \leq 2^n,$ so we've reduced the inequality to showing that $2 \leq 2^n,$ but $n \geq 1,$ so $2^n \geq 2^1 = 2.$

[vital dew]

EGA 1:

[balmy nacelle]

Can someone check that I have the right idea

so you got

Where exactly can you tell where the internal nodes end?

it's not just B and C right?

[pale epoch]

what's an internal node

[balmy nacelle]

it's ok I got it

different word for parent node

[weary tiger]

anyone know how to prove that

assume its not and there are 2 numbers that give the same remainder mod M

[weary tiger]

i don't understand what you mean

[faint narwhal]

Look at the definition of injective

[white monolith]

i have question

is anyone online

[weary tiger]

[weary tiger]

lmfao how do i answer this

[white monolith]

Help

Question 7

[gleaming zephyr]

= here means logicall equivalent

try using P <--> Q = P --> Q and Q-->

and using P--->Q = not P or Q

and demorgen

@white monolith the algebra should work out ig

[white monolith]

Ty

Question 8?

[gleaming zephyr]

try induction

strong

@white monolith

[left thistle]

Can I get help with a few homework questions? I’m kinda stuck on them

oki

@neon thorn you said post the questions?

Yeah I tried the problems but I don't know how to finish them

Like for 5 and 6 I can try setting up the problem, but I'm stuck on what to do next

[left thistle]

🤔

When you say it like that, it's easier to break down that I managed to solve it. Thxs for the help man I really appreciate it

[patent rover]

we have k choices for a, then 100-k choices for b and then 100-k choices for c

which is k(100-k)^2 choices

but the solution says this:

could anyone help me understand this more intuitively

[stray reef]

what's X

[mellow girder]

For 1 i got C(6/1)87*C(5/4)*8^4

Sorry, it's

C(6/1) * 8 * 7 * C(5/4) * 8^4

for 2) i got

C(6/1) * 4 * 3 * C(5/4) * 8^4

Is this right?

[lucid fog]

hi, can someone help me with 2 proof questions

?

[lucid fog]

im not sure how to do 4 and 5

or how to start it

[weary tiger]

@lucid fog I'm sure there is a more elegant solution to this (probably with some theorem like whatever the answer to q3 was) but you could try brute force contradiction for each of $X_k \equiv \pmod{0 ,1,3,4}$

[vital dew]

Botnuke:

[weary tiger]

and q4 implies q5 very quickly

the cases for 1 and 3 mod 5 are easy, and I think I have a contradiction for 0

I will try 4 mod 5 as well

[lucid fog]

I'm not sure what you mean by brute force each of Xk because they dont give you values for what k is

I was planing to prove num 5 first because every 3^power has a pattern

[weary tiger]

oops I mistyped

$X_k \equiv 0,1,3,4 \pmod{5}$

rule out these possibilities

and it must be congruent to 2 mod 5

wtf

[lucid fog]

ye

[vital dew]

Botnuke:

[weary tiger]

proving either of q4 or q5 will give the other

but q5 says more than q4

I don't think it could possibly be easier

[weary tiger]

I think I've solved it btw but not 100% confident in my proof

I (maybe successfully) used well ordering on the 0 and 4 mod 5 cases to show there is no least X_k which is congruent 0 or 4 mod 5

which also implies q5

oh wait I'm dumb

there is more to it for 1 and 3 cases

ok I used well order for those as well

[stray reef]

this does not count the passwords of the kind you are asked for

10 * 36^5 counts the passwords which begin with a digit

[weary tiger]

Hey if i can get some help with this I’d appreciate it

<@&286206848099549185>

[weary tiger]

i need help ive tried 5 times and still getting wrong

[tardy jacinth]

not the right channel

[spark oyster]

i do wonder why discrete math attracts so many questions which absolutely have nothing to do with discrete math

do people just come here if they want their homework solved discreetly

[steady tendon]

help please

[pale epoch]

i do wonder why discrete math attracts so many questions which absolutely have nothing to do with discrete math
@spark oyster it's because classes like "intro proofs" or even just "intro mathematics" are labeled "discrete mathematics"

[spark oyster]

huh, weird

[pale epoch]

it's often just math for computer scientists or sth

[stray reef]

@steady tendon too vague.

what are we coloring? vertices, edges, perhaps both? how many colors are available? what conditions must the coloring satisfy, if any?

[steady tendon]

vertex colouring, such that no two adjacents can be of the same colour

[stray reef]

how many colors are available?

[steady tendon]

all you can

7 in that case

7 nodes

[halcyon ledge]

Well just one, the vertex in the middle has 6 neighbours, so you need a different color for each of them

[near prawn]

the first one looks very wrong

why are you adding them

[lyric pumice]

@pale epoch Maybe it's also because they see the word "math" in the channel name.

And all the other channel names look more specific.

[lilac gazelle]

[sacred furnace]

Could I possibly get some help on this problem?

[gleaming zephyr]

what do u need help with

what is it that u do not understand

[sacred furnace]

How to preform the function for (1,1)

Like (1,1)G(1,1) = 1-1=1-1?

Would the 5 elements be 1,1,1,1,0?

[gleaming zephyr]

we say (a,b) R (c,d) iff a-b = c-d

so u want the equivalence class of (1,1)

[(1,1)] = {(a,b) | (1,1) R (a,b)} = {(a,b) | a-b=0}

[sacred furnace]

Is that a modulo?

[gleaming zephyr]

no

[sacred furnace]

So if a-b=0 then 0=c-d ?

[(1,1)] = {(c,d) | (1,1) R (c,d)} = {(c,d) | c-d=0} ?

[sacred furnace]

I'm still lost on this problem. I can't find anything online or in my book

[sacred furnace]

Do I want a-b=c-d to be 0=0 or 1=1 for the elements of [(1,1)] ?

[weary tiger]

imma guess the latter (1=1)

[sacred furnace]

Ya that's what I'm thinking

[quasi finch]

any1 wanna help me out real quick? try not to give me an answer but give me some methods. I missed the first half of this lecture and my professor doesnt have online notes..... so i missed this "easy " part of lecture 🙂 please and ty 🙂

<@&286206848099549185>

[hexed granite]

which problem?

[quasi finch]

problem 1 i think i got problem 2

[hexed granite]

so for 1a, you have 26 letters and 2 cases so 26*2 = 52 and 10 numbers so 52 + 10 = 62

and you have lengths 9 - 11

so for the first 9 characters, you can choose any of the 62 characters

and for the 10-11 you have the option not to choose any so that is 63 options

but for the 11th characters you only have that choice if you chose a character on character 10

[quasi finch]

uhhhhhh....

what if one wants to not have case and or numbers and just all letters?

or is that not how it works

or is that possibility ^ included in these options

[hexed granite]

its included

[quasi finch]

no idea man.....

[idle violet]

any ppl reading discrete math

[lyric pumice]

Yes.

[patent moat]

How many possible combinations in 10 character password?
using only small letters
it's 26^10 right?

[halcyon ledge]

yes assuming you're allowed to use any letter as many times as you want

[patent moat]

ok then it becomes a bit tricky and im stuck

it asks lower case only + no repitition + in alphanumerical order

e.g bdgjkmpqr

[halcyon ledge]

well no repetition should be simple

[patent moat]

yes

[halcyon ledge]

alphanumeric makes hard

[patent moat]

26* .... * 17

yea exactly.

[halcyon ledge]

exactlly

[patent moat]

but the alpha order it's like i need to form a formula

[halcyon ledge]

hmm okay

I would go about it the same way as with the no repetition only now include the additional restriction

so say c is the first letter, then for the next position you only have 24 instead of 25 and so on

[patent moat]

so if we say we picked the 15 letter of alphabet the second one needs to be in the range [16,26]

[halcyon ledge]

exactly

and from there you walk to the next until you hit the end

[patent moat]

N * (N-1) * ....(N-10) ?

nvm that's stupid lol

[halcyon ledge]

hmm I don't think there is a closed formula

[patent moat]

so i should go about it with a reference starting point ?

[halcyon ledge]

because if you start with q you only have one possible password

qrstuvwxyz

so the amount of possibilities always depends on what letters you place

[patent moat]

so u suggest starting from a reference point ?

and that possibiliti always changes tbh
if u start with A then u have 25^10 possibilities but then if u pick Y u have 1^10

[last timber]

look

for the first letter you have only 16 choices

since if you take for exampe w as first letter no password is possible

[patent moat]

true.

what about the second , there are too many options

[last timber]

no, why you always have to preseve 10-k letters that are bigger lexicographically then the k-th letter

[patent moat]

i think the solution is 16 * 15* ..... * 7

[last timber]

and let us always thing that k-th letter we take is always the biggest lexicographically

[halcyon ledge]

No 16 * 15 * ... * 7 would include illegal strings

because if you place d as the second letter you have less options

[patent moat]

so if u pick K as the first letter that takes from [16,26]

ehh im burned.

i think im missing something can't be that hard.

[halcyon ledge]

whats the course?

[patent moat]

discret math

[halcyon ledge]

as in what tools did you get to solve the problem?

[last timber]

or look abcdefgh

[patent moat]

combinational

combinatorics

[last timber]

wait i am idiot

lexigoraphical order

[patent moat]

yea e.g bdgjkmpqr

[last timber]

that means it is just combinations of 10 elements without repetition out of 26

since {a,b} and {b,a} is in set notation equal

[patent moat]

that was the previews question

which is 26 * . ... * 17

[last timber]

yep

well you can check it in the way you have password "abcdefgh" and you have 17 options

then "abcdefgi" and you have 16

and so on

[patent moat]

arent those the same

anyway

it asks using only letters A and Z
is this
2^10 * 10

[last timber]

no

since here you can have aaaaaaaa

[runic cedar]

is the coefficient of x^12 in (1-x)^-7 0?

because I'm getting 7C12 which is a no no 🤔

[last timber]

@runic cedar use mclaurin series i think

or probably stirlings numbers

[patent moat]

yes u can have aaaaaaaaa

or bbbbbbbbb

so each time u have 2 options a or b

2^10 * 2^10 .... 2^10 , 10 times

no ?

[last timber]

wait why

you said repetitions is not allowed

[patent moat]

this is another

10 char password if we use only letters A and Z

[last timber]

A and Z?

[patent moat]

ye

[last timber]

then yep, 2^10

[patent moat]

what if

[last timber]

not 2^10 2^10

just 2^10

[patent moat]

2^10 * 10

[last timber]

no

look

a or b - two options

a or b for each of them next

[patent moat]

ah ok

yea

[last timber]

a or b for each of before

and soo n

[patent moat]

so each

u have 2 options

2^1

for 10 letters

so

2^1 * 2^1 * 2^1 .... * 2^1

oh yea should be correct

what if , use only the letters A and Z but same quantity for each.

so it could be
AA BB AA BB AB

or
aaaaa bbbbb

[last timber]

yep

as my professor tells it is not important which is the letter

it could be even spongebob

[patent moat]

this is another one

what if , use only the letters A and Z but same quantity for each.

for 10 char pass

[last timber]

wait wut

like only AAAAAAAA or ZZZZZZZZZZZ

[patent moat]

not really

[last timber]

obviously 2 ways

[patent moat]

it could be AAAAA BBBBBB

5 a and 5 b

or

AA BB AA BB AB

or

[last timber]

oh i see.

[patent moat]

AB AB AB AB AB

[last timber]

it is arrangement i think

10!/5!5! anyway

[patent moat]

why

[last timber]

well, since a and b are equal in quantity then it could be only 5 of each

[patent moat]

true

[last timber]

and then you just arrange all of A and B into two "boxes"

[patent moat]

wum

[last timber]

like here

[patent moat]

10! / 5! * 5! , 10 goes for 10 char long , 5 goes for each A and 5 for each B

hmm ok

2 tricky onesl eft.

but i have no idea what to do

use only small letters + digits , no repitition + take turns one letter one digit or the other way

e.g 1a2b3j8g5l or k3j5s0l1m9

So we have
26 for the first one + 10 for the digits

i stuck on those because how u gonna solve it , it always depends if the first is 36 then if we pick number second is 26 if we pick letter second is 10

[last timber]

actually your logic is right

since multiplication is commutative

you have 36 ways for the first pair

or smthng like that

sec

10*26 for the first

[patent moat]

For the first one yes but second one really depends on first.

[last timber]

9*25 for the second pair

[patent moat]

no wait

if first is letter second needs to be digit

and the otherr way around

[last timber]

yes, but look

we take them by pairs

a1a1a1 for example differs from 1a1a1a only by order

so we can do like that

[patent moat]

not only by order

letters are 26 , digits are 10

[last timber]

we count all ways when the first is digit and then multiply by 2

letters are 26 , digits are 10
and?

2610=1026

26x10=10x26

[patent moat]

26x10 is the possible outcomes if we pick a digit or ch

char right?

ok yea true makes sense.

[last timber]

look. let us suppose we use order char - digit

[patent moat]

so 2610 for first
2510 for second
24*10 for third?

[last timber]

yes

and so on until five pairs are done

then you multiply it by two since order digit-char also allowed

[patent moat]

i see

so it's

(26* 10 .. ..... 17*10 ) *2

[last timber]

why 10?

[patent moat]

that's what i was thinking now

[last timber]

26x10x25x9x....x20x5x2

[patent moat]

shouldnt be

until 22 * 6

not *20 * 5?

[last timber]

oh yes

until 22

and 6

[patent moat]

ok and do u know how to do small letter + no repitition + lexigraphical order?

e.g bdgjkmpqr

thanks for ur help btwi truly appreciate it

[last timber]

ok and do u know how to do small letter + no repitition + lexigraphical order?
amm

we did it just before no?

[patent moat]

no we never figured it out

[last timber]

that means it is just combinations of 10 elements without repetition out of 26
@last timber

[patent moat]

hmm

the thing is that's the b part

asks exactly that

small letter no repittion

and the c part asks
small letter + no repittion + in order

so im not sure if its'; the same.

for first we have 26
for second we need to know the first.

[last timber]

it should be the same i suppose

oh, no

maybe