help-49

[wraith dirge]

wdym

[last slate]

dude like

[coarse wing]

[lavish venture]

what?

[wise flicker]

There are 18 slots, choose 4 C(18,4) and then arrange A729 4!

[wise flicker]

Timezone

[last slate]

p(32,14) is the set of 32 characters in premitations in the form S = {32}

So then S x S-1 x S-2 x … x S-31

[lavish venture]

whats it to you

[wise flicker]

:pepe_cry

[coarse wing]

alr ima take a nap

[last slate]

4! * 17?

[wise flicker]

Is it?

[last slate]

I genuinely hate this

[lavish venture]

why major in stem then

[last slate]

Bro what do they want from me

[lavish venture]

or are you doing some sort of other science

[last slate]

its just this particular topic of counting

[coarse wing]

is this breakdown

[last slate]

nah

im gonna continue majoring in stem

[wise flicker]

Take a nap

[coarse wing]

@prime hornet !!!!!!

help him

[prime hornet]

||honestly, I hate this sort of counting too ||

[last slate]

exactly

[coarse wing]

my brain overloaded

[last slate]

this is universal bruh

[hallow crane]

Try chemistry

[last slate]

i love math but this tests your limits

[coarse wing]

shall we back to the donut 🍩

[last slate]

id rather do physics every hour of my life than this

[coarse wing]

try physics

[last slate]

yes

[wise flicker]

nick crosses it

[last slate]

Its an option

[prime hornet]

I'm not good enough at these problems to do that :c

[coarse wing]

genuinely i can solve these but im kinda bad at explaining

yeah

f#$^ counting

[wise flicker]

Ts how pnc is

[last slate]

lowkey i wanna hop on vc for this

[hallow crane]

Invariance, monovariance,etc are far superior forms of combi

[wise flicker]

Take him @hallow crane

[coarse wing]

heres the previous topic

[hallow crane]

Wdym?

[wise flicker]

Vc

[wraith dirge]

i hate zybooks sm

[hallow crane]

Nah i would skip

[wise flicker]

Last night

[coarse wing]

i told him to do the bars and stars

[wise flicker]

Lmao why

Its repeated letters

[coarse wing]

i mean this probably the simple approach for me i think

$\binom{8 + 3 - 1}{3 - 1} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = \mathbf{45}$

[grand pond]

linh

[coarse wing]

leon

think

[wise flicker]

Hmm.. usually we jus divide by no. of repeated

[coarse wing]

read the chatt5

aight

cya

[wise flicker]

@leon you there

[last slate]

yes

[wise flicker]

K then 4!×C(18,4)

What is it

[last slate]

i dint understand what you did there

[coarse wing]

[last slate]

there are 4 letters that must exist in the mix,

just calculate how much you can rearrange them

[coarse wing]

let me turn this channel notification off

[last slate]

which is 4!

And then

idk

[wise flicker]

Length of password was 18, we choose 4 spots for those letters then arrange at those 4 spots

Total spots were 18 so C(18,4)

[last slate]

whats C (18,4)

[wise flicker]

Combinations

[last slate]

Why bruh

what is the goal here

[wise flicker]

To choose 4 spots

[wise flicker]

Understand this?

[last slate]

no

[wise flicker]

When you multiply C (18,4) * 4!

[last slate]

differential equations is easier than this

[wise flicker]

You get P(18,4)

[last slate]

ye?

[coarse wing]

@last slate do you truly understand how permutations and combinations works and its theory.

[wise flicker]

Lmao

[last slate]

i get how premutations work for sure, combination is just permutation but you map r! to a single subset for the duplicated ones

[coarse wing]

because if you dont get it you cant understand what we are explaining

[last slate]

im not even tryna be funny it actually is

[wise flicker]

Whose video you sent em?

[coarse wing]

scroll the chat

[wise flicker]

Leon

[last slate]

aw man

yes

[wise flicker]

You got answer

[last slate]

no

[coarse wing]

if its too hard or you need to revise you can talk to your professor

[last slate]

my professor doesnt explain well

ive been self studying my entire career

Except for one class where the professor gave a shit and had some conscience

[coarse wing]

http://www.youtube.com/watch?v=XJnIdRXUi7A

luckily i saved this in my playlist

[wise flicker]

Crazy was gonna shareda same

[coarse wing]

im just faster

😆

[coarse wing]

@last slate this video probably helps you in self studying

[last slate]

alright thanks, ill check tbis one out

But like

can i still get an explination for that question

th

This

what does P(18,4) do

[coarse wing]

check the video first

eventually you will found out the explanation by yourself

[last slate]

okay

thank you

[coarse wing]

yw

[last slate]

okay i watched it

but

i already know what it means to do a permutation and combination

but my question was why do a permutation like that one in the first place

why

im saying why

cooked

<@&286206848099549185>

[void lichen]

are you asking why you need to do a permutation here

this is your original problem right

[last slate]

whats the set of 18

[last slate]

yes

[last slate]

specifically P(18,4)

[void lichen]

18 is the length of the password

which is given

[last slate]

why are we considering the lenght

[void lichen]

im pasting the problem again for my reference

each valid password

will have to be 18 characters long

now, the restriction states that it must contain a, 7, 2, and 9

[last slate]

yes

[void lichen]

now let us look at what the first character of our password can be

it can be any of the 4 characters

[last slate]

yea

[void lichen]

we must

then the second characrter can be any fo the 3

because we cannot repeat characters

so we have 4, 3, 2, 1 which is 4!

the first character can be a, 7, 2, 9
Suppose we pick a
then second can be 7, 2, 9
then we pick 9
third can be 7, 2
we pick 2
fourth has to be 7

[last slate]

yeah here youre talkin about arranging them in a slot of 4

?

i mean 4 slots

[void lichen]

yes of the 18 slots we want four to be the MUST CONTAIN characters

rather

the first character can go in any of the 18 slots

second can go in any of the 17 slots

third can go in any fo the 16 slots

last can go in any fo the 15 slots

now these four characters can be arranged among themsleves in 4! ways

[last slate]

And those 4! Ways can be in different ways in the 18 slots

is that what youre saying

[void lichen]

yes

[last slate]

okay i agree

[void lichen]

so we have 18 then 17 then 16 then 15

and then these four characters can be arranged in 4! ways

okay now what is a permutation it is n! upon (n-r)!

[last slate]

You lost me here

[void lichen]

the first character suppose 'a' can go into any fo the 18 places

the second character '2' can go in any of the 17 places

because a took up one place

accordingly, the third character '7' can go in any of the 16 places

now, the fourth character '9' can go in any fo the 15 places

the answer is for why it's 18 p 4 is understanding it like 18 times 17 times 16 times15

first goes into 18 slots

second 17 slots

third 16 slots

fourth 15 slots

which means 18 times 17 times 16 times 15

[last slate]

Yes but how does the permutation look like for that

[void lichen]

okay so

a permutation is

npr which is n! upon r!

here n = 18

and r = 4

so 18! upon (18-4)! would be

18! upon 14!

which is actually just 18 times 17 times 16 times 15

[last slate]

but youre doing the premutation on the 18

which is like what an empty set?

or what

like

what are you putting in slots of 4

[void lichen]

we are putting the four characters in 18 slots then 17 slots then 16 slots then 15 slots

so we have picked

4 slots

do you get that

[last slate]

yes

i just dont get how P(18, 4) solves it

[void lichen]

p(n, r) calculates the number of unique arrangements possible when choosing r items from a set of n items

[last slate]

you’re doing a permutation from the set of 18

yes 18 what

what 18 items are we taking

[void lichen]

18 places

[last slate]

so places

okay

[void lichen]

18 different places where can put the characters in

out of which we just want to put the characters into 4 palces

[last slate]

sure okay

and then in each place the letters can be arranged 4! Times

So thats 4! * P(18, 4) no?

[void lichen]

p(18, 4) already accounts for

that

[last slate]

nevermind it aleady counts for it

[last slate]

ue

[void lichen]

yup

[last slate]

ye

[void lichen]

think of it less in terms of the rigid formula npr

[last slate]

well thanks man thats all i needed to know

🤣

[void lichen]

cool

[coarse wing]

im back

[last slate]

hi

yeah i just needed to know wth 18 was

[coarse wing]

okay what about the donuts

[void lichen]

was i able to explain properly

[last slate]

yeah thanks

[void lichen]

i didn't know how to explain the fundamental idea of permutations

cool alr

[coarse wing]

@last slate you are missing the donuts

[last slate]

now time for subsets 😈😈😈

[coarse wing]

no

[coarse wing]

okay if this problem already end just .close

it

[last slate]

not quite

[last slate]

which

the one i already did

[coarse wing]

type .close

[last slate]

rn?

Sure

.close

[midnight plank]

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[coarse wing]

bai bai

[midnight plank]

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[fossil wagon]

anyone know how to do qn23 b and c?

[paper mantle]

how would u find the distance between A and B?

[fossil wagon]

that would be the magnitude of AB

which is 20.6

[paper mantle]

how do u express that mathematically?

[fossil wagon]

the formula for line segment

[paper mantle]

okay

so for ur B, ur question says AC = 2/5 AB

u have ur AB so u can find ur AC

[fossil wagon]

yeah I've gotten 8.246

[wooden badger]

Section formula

[fossil wagon]

but how would I find two unknowns x and y coordinates?

[wooden badger]

You've been given the ratio of AB and AC

[fossil wagon]

but isnt that to find the length segment

not the coordinates?

[wooden badger]

Do yk the section formula?

[fossil wagon]

yeah

d = √[(x₂ - x₁)² + (y₂ - y₁)²] this right?

[wooden badger]

So you could assume the coordinates as variables and apply tht

[wooden badger]

Thts distance fornula

https://share.google/OG7rm2sa4VueM8RP3

[fossil wagon]

oh then idk whats section formula

[wise flicker]

Ts section formula

[wooden badger]

...?

[wise flicker]

They asked section formula

[wooden badger]

Oh

[midnight plank]

@fossil wagon Has your question been resolved?

[midnight plank]

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[midnight plank]

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[midnight plank]

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[severe grail]

I need some help

[severe grail]

what is it asking?

[obsidian anvil]

its asking you to use the given mean and standard deviation to calculate exam scores

for example if the average (mean) score for an exam was a 70 and the standard deviation was 8, the score that is one standard deviation from the mean is the mean + 1*std (standard deviation)

the score that is 1 standard devation below the mean is mean - 1*std

does that make sense

[severe grail]

do we add 1 to z and solve the equation

for the first question

[obsidian anvil]

the expression you would be using is

$\mu \pm \sigma$

[grand pond]

uni

[obsidian anvil]

where mu is the mean and sigma is the standard deviation

[severe grail]

is it ok to use x bar?

[obsidian anvil]

If a score is one standard deviation above the mean, you add one standard deviation to the mean

[obsidian anvil]

sure but it doesnt really matter

you dont need to necessarily use variables to answer this question

[obsidian anvil]

If a score is two standard deviations below the mean, you subtract 2 standard deviations from the mean

[severe grail]

ohhh ok I think i get it now

basically

a higher z score means

more area on the curve

[hushed mauve]

a higher z-score means you are further along the curve (as in further to the right)

you can only have an area if you're calculating P(x < or > z-score)

[severe grail]

P(x < or > z-score)

I dont understand this

[hushed mauve]

I thought I made it simple enough

basically, if you're looking for a range of z-scores (like P(x > 1.5)) then yes, you will get an area

otherwise you're only getting one point

[severe grail]

did u mean P(z>1.5)

because thats what is written in some of my problems

[hushed mauve]

just a var difference, but sure

[midnight plank]

@severe grail Has your question been resolved?

[midnight plank]

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[wind oxide]

consider a $7$ digit number $abcdefg$ such that $d$ is the greatest digit and digits towards the left and right of $d$ are in decreasing order (from left to right). then the total number of such numbers in which all digits are distinct

[grand pond]

rak³en

[wind oxide]

wait a minute

9c7 * uhh

36?

hmm does not seem to match

[slow thorn]

i think its like 1234321

oh from left to righ

mb

so like 3214321?

[wind oxide]

its alr

[wind oxide]

but all distinct too

yes btw thats what i am thinking

[wind oxide]

okay i understand why this is wrong

OH

[formal orchid]

This is very similar to that AMC 10/12 problem.

[wind oxide]

9C7 6C3

IT MAKES SO MUCH SENSE NOW

[hallow basin]

Lmao I have done this one b4 lol

[wind oxide]

.close

[midnight plank]

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[hallow basin]

uh

already?

that was fast

[formal orchid]

Oki, so $7 \le d \le 9$

[wind oxide]

i am done

🙂

[formal orchid]

DW

[wind oxide]

this was my initial approach btw but i saw the options and knew there was some pncier way

[grand pond]

Erebus

[slow thorn]

arent you like excluding zero here

[formal orchid]

That seems to be the case.

[slow thorn]

getting the same answer considering from 1 to 9

cant you just safely say 10C7 times 6C3

[formal orchid]

||10C7 * 6C3 - 9C6 * 5C2|| right?

[slow thorn]

why so

zero would never come to the first position

[midnight plank]

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[formal orchid]

||10C7 * 6C3|| is for all possible cases (including 0)
||9C6 * 5C2|| is for cases where 0 comes first (excluding 0)

[formal orchid]

.close

[midnight plank]

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[slow thorn]

why would zero come first

arent the numbers arranged in descending orderr

[formal orchid]

I mean I'm using complimentary counting.

[slow thorn]

yeah but the digits are already said to be arranged in descendign order

[formal orchid]

1239876

Is the above one not of such possibilities?

[slow thorn]

nah i think 3219876

[formal orchid]

Ahh oki

Then ||10C7 * 6C3|| is valid

Since 0 can't be first

(misread the question 😅 )

[midnight plank]

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[twilit field]

Just wanted to make sure I'm not tripping

all I do is find |J|

and set x= ( 3v-u)/10

where J=1/10

so the joint pdf id (3v-u)/(100)?

[zealous schooner]

What

What is J

[twilit field]

The jacobian

this is my formula

[wooden badger]

Ping helpers?

@twilit field

[twilit field]

yea?

[twilit field]

Like just want to be sure the domain doesnt affect the method, or how its applied right

[twilit field]

this method

[marble hedge]

Joint Density

Damn
i don't know about this.

[twilit field]

.close

[midnight plank]

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[odd totem]

Ho can i draw it correctly ?

[lusty python]

Draw what correctly?

[gusty falcon]

part b lol

[lusty python]

you're not the OP

[odd totem]

Who I draw A the symmetric of l with respect o and the others

[gusty falcon]

OP's question was clear enough, so i saved him the trouble of having to rewrite the same thing again

[lusty python]

Im thinking if he can't draw a right triangle then im smashing my device

[wise flicker]

chill, let's focus on op

[lusty python]

Symmetric should be smth like O is the midpoint of AL? Im asking for clarity since im not familiar with geometry with english

!noans

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[lusty python]

Ig im right from the drawing someone deleted

[marble hedge]

Sorry

[gusty falcon]

bruh 😭

(just google it)

[marble hedge]

Finding the third side would be extremely difficult but possible

If that was true

[lusty python]

Basically making point B the midpoint 🙏

[midnight plank]

@odd totem Has your question been resolved?

[lusty python]

So in this diagram, you can extend LO such that O's the midpoint of AL

[midnight plank]

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[thorny folio]

Hello great math fellows, I need some hint regarding a numerical analysis question I have.
I'm totally lost, and confused.
Here is the question:

Considering the following iterative methods.
\begin{flalign}
x_{n+1} = z_{n+1} - \frac{f(z_{n+1})}{f'(x_n)}
\qquad\qquad
z_{n+1} = x_{n} - \frac{f(x_n)}{f'(x_n)}
\end{flalign}
Assume $f(p) = 0$ ($p$ is a root of $f$). Prove that the sequence ${x_n}$ converges to $p$ with convergency order of $3$.

I'm totally lost. The iterations look like Newton's method, but it seems one is dependent to another one. Shall I just plugin $z_{n+1}$ to the first one, and try to solve it?

In my text books states that the Newton's method has order of convergency of 2 if $f$ is linear, otherwise it's 1.

Thank you for your time and consideration.

[grand pond]

idiot_max

[midnight plank]

@thorny folio Has your question been resolved?

[runic hamlet]

you can try replicating the proof in your textbook of the order of convergence for newtons method

although your claim about that seems false

if f is linear then it should converge immediately

and it should have quadratic convergence quite often

but you do need more assumptions for your function f

[thorny folio]

Yes of course, my bad. Sorry I'm very tired and confused.

[thorny folio]

Thanks.
To prove that the $f$ has convergency order of $3$, shall I just prove that $\alpha=3$ in the following statement?
\begin{flalign*}
\frac{\left|x_{n+1} - p\right|}{\left|x_{n} - p\right|^{\alpha}} = c
\end{flalign*}

I think I have confused myself with Newton method, fixed-point, and the above statement.

[grand pond]

idiot_max

[runic hamlet]

yes

<= would be enough I suppose

[thorny folio]

Much appreciate it. And any hint how to find the constant $c$ (asymptotic error const) ? and is it legal to assume $p=0$ ?

[grand pond]

idiot_max

[runic hamlet]

well the classic way would be to plug in a few terms from the taylor series and work with that

I think

so try that

look at how your textbook proved the claim for newton

[thorny folio]

Thank you. Yes, understood.

[thorny folio]

Thank you so much sir, really really appreciate your time and consideration. Math bless us all.🍻🤓🤘

.close

[midnight plank]

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[wispy pawn]

.open

hi

[wise flicker]

Yes

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[scenic wyvern]

!occupied unfortunately. if you have a question, please grab an available channel

[midnight plank]

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[dusty portal]

gggggg LALALAL (supposed to be a really dramatic entrance)

[dusty portal]

So I'm completely lost on the 2nd part of this proof...

I intuitively understand how it works, but I have no words for it

[latent wadi]

Maybe you should send us a photo of it

[dusty portal]

wdym

I understand it like how P(A cup B)=P(A)+P(B)-P(A and B)

Waaaaaaaaaait

Lemme draw a probablity diagram rq, just continue with what you were saying

[dawn dagger]

of their intuition?

[latent wadi]

It shouldn't be too bad if they already have the intuition for the problem

[dusty portal]

ouuuuu

[dusty portal]

Well this is it

[runic hamlet]

pick a basis

[dusty portal]

{e1, e2} ig

For both A and B?

[runic hamlet]

ok how did you show the first part...

[dusty portal]

Ok I might be screwed

brb, lemme get my book

goddamnit

[prime hornet]

<@&268886789983436800>

[prime hornet]

pick a basis for A + B

[dusty portal]

Oh

I keep on seperating these questions vro 🥀

[dusty portal]

Picked basis for A and B, combined, then showed that the set spans A+B

Ok I think I see what I can do here

And I choose basis for A cap B right?

aight

I'll see if I have any other problems with this

.close

[midnight plank]

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[midnight plank]

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[sturdy barn]

I am so lost how to do this

[sturdy barn]

So I think A

Just write the formula and add the numbers?

[sturdy barn]

W bot lol

For 4a. I did

€C=275x1+1+150+25n

Is that good?

[wooden badger]

Where is c and w?

[wooden badger]

I mean w

[sturdy barn]

Oh

Uh

[sturdy barn]

Oh so 1 is suppose to be 1w?

[wooden badger]

Acc to the formula yes

[sturdy barn]

€C=275x1w+150+25n

[wooden badger]

Yesyes

[sturdy barn]

Okay so now for 4b uhhh

[wooden badger]

Tht i believe would be correct

Just insert values

[sturdy barn]

I think so too

[wooden badger]

U hv this fornula

[sturdy barn]

Four people=100n?

[wooden badger]

N is the no. Of ppl

So yes

[sturdy barn]

€C=275x2w+150+100n?

[sturdy barn]

And then I have to work out the value of c?

[wooden badger]

Yup

[sturdy barn]

Lowkey forgot but I will try

[wooden badger]

Forgot what its js calculation TT

[sturdy barn]

I need to serpate the w's and the n's on different sides?

[wooden badger]

You just need to plug in the values

And calculate

[outer saffron]

For question b they provided the w's and the n's so you just substitute them

[sturdy barn]

Ohhhhhhh now I know how to do these thanks

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[last slate]

Hi i need help understanding the logic behind this formula

[wooden badger]

Thte ncr?

[quiet grove]

go through these

[last slate]

what?

[quiet grove]

its called the stars and bars method u can check it out on yt

[last slate]

i did but they don’t explain it properly

i just need someone to explain to me the intuition

[quiet grove]

read this once

[pearl mist]

use it when you want to distribute identical objects in distinct boxes or people

[wooden badger]

Ooh the gap method

[last slate]

yeah but it finds what

the amount of what

and does it count duplicates

[pearl mist]

number of ways to distribute identical objects among distinct people

maybe thats why i learnt it as beggars method

[quiet grove]

yeah

[last slate]

say you had a variety of four flavored cupcakes

You would like distribute it amongst 12 people?

so you would have like 3 vanilla 2 glaze 4 crums and 2 chocolate

and you would see how many ways you can make a similar combination?

but then what about duplicates n stuff

[quiet grove]

this wouldnt be applicable here because the cupcakes are distinct

neither are the people so u cant manipulate it in the opposite way

[pearl mist]

well i suppose you dont distribute people among cupcakes

you need identical cupcakes and different people

[last slate]

is this the whole “distinguishable” vs indistinguishable thing

for God’s sake explain this to me man ive been trying to understand that too

@quiet grove

[quiet grove]

if 2 things are the same they are undistinguishable

2 people are not the same so they are distinguishable

[wise flicker]

Stars and bars applies directly only to identical objects

[quiet grove]

a vanilla and a glaze cupcake are not same so they are distinguishable

[pearl mist]

look ill tell you this the way i learnt suppose there are 10 beggars well each person is different and you want to give them money say 5 bucks each can you distinguish each set of 5 bucks for all the beggars? no bc money is money but can you distinguish the beggars? yes

ok forget about 5 bucks say 1 coin for each thats more easy to understand you cant distinguish 2 coins if they have the same value

[quiet grove]

yeah

[wise flicker]

You can do this way, how many ways can you cholse flavour for 1 person

[last slate]

yeah and

and what are we counting for

what multisets are we counting here

[quiet grove]

just like the number of solutions for x1+x2=4

0,4
4,0
1,3
3,1
2,2

this is what you get from the formula

[pearl mist]

number of ways you can give 1 coin to the beggars

[quiet grove]

(4+2-1)C(2-1)=5C1=5

[quiet grove]

and we got 5 cases here as well

[last slate]

okay wait how many beggars do we have as an example?

just as an example like say 5

okay and they are distinguishable

now what

[pearl mist]

ok so there are 5 beggars and you have to give each one of them one coin so you should atleast have 5 coins ryt

[last slate]

yes

right

[pearl mist]

so here m and n both are 5; 5 coins and 5 beggars

[last slate]

And we are trying to find what multisets exactly

[pearl mist]

multisets as in?

[last slate]

like sets with repeated objects

the whole section is titled “counting multisets”

[quiet grove]

{1,2,2,0,0} {1,1,1,1,1} {1,0,3,1,0}... etc

[last slate]

each slot being a person?

[quiet grove]

yaah

[last slate]

wait so not all of them get money

[quiet grove]

and the number is number of coins

[quiet grove]

yeah

because the formula u sent specifies there are no restrictions on how much can go to each person

[last slate]

okay so can you please explain to me after this step why we get to that specific formula to calculate all these possibilities?

[quiet grove]

if u want all of them to get at least 1 coin the formula would instead be (n-1)C(r-1)

[pearl mist]

not necessarily the formula still works if you have 5 beggars and 4 coins whatever may be the number of ways one beggar will not get a coin

[quiet grove]

i sent u a pic with the explanation read that first nd lmk if u get it

its basically the number of ways of arranging n objects and r-1 partitions

imagine u have n objects lying on the ground

if you want to make r groups out of them, you have to place r-1 partitions between them

[last slate]

but there isnt any paritions when i rearranging the money to the beggars

we just give the money

in different possible ways

[quiet grove]

do u want a beggar specific derivation 😭

[wise flicker]

Lmao beggars act like slots

[pearl mist]

[last slate]

not necessarily

[noble lance]

do you understand how the beggers and coins correspond to the "objects" and "varieties" from the question

[last slate]

okay yeah but there are no seperations

[wise flicker]

There's a reason ts famous as beggar's method

[last slate]

i guess so the money is the variety

[quiet grove]

we love beggars

[last slate]

and the objects is the beggars

[wise flicker]

Put no. of beggars-1 separations

[pearl mist]

ts looks ugly

[last slate]

but why

[noble lance]

I think it's the other way around? because each beggar can get multiple coins
and in the question, each variety can yield multiple objects

the correspondence has to make sense to you first, otherwise understanding the beggar variation will not make the original any clearer

[last slate]

i don’t understand bro can we do a different analogy where the numbers arent the same

like the cupcake one

you have 4 varieties and you wanna give to 12 people

[quiet grove]

because if you draw 2 lines, it separates the space into 3 parts

so we're drawing 4 lines, which would divide the space into 5 parts, and you put a beggar in each part

[last slate]

so you wanna count the ways that all 12 peoples would get cupcakes

[noble lance]

okay, just quickly, I saw stars and bars mentioned earlier, do you know what that is or have you never seen it before?

[quiet grove]

if you want ALL the people to get at least 1, theres a separate formula for that

[last slate]

ive seen it but its very confusing

[last slate]

i see, can we please start with were its allowed for one of them to not get a cupcake

[noble lance]

okay, that's good, are you comfortable running with stars and bars? that's the quintessential example for this

[noble lance]

let's say our n objects are stars. we have all of them in a line, but each may come from a different variety. let's group them into the m possible varieties, and separate each group by a bar.

suppose that ends up looking something like ★ ★ ★ ★ | ★ | ★ ★

with me so far?

[last slate]

maybe

[last slate]

yes

[noble lance]

okay great

in this case (★ ★ ★ ★ | ★ | ★ ★) there's clearly 7 stars (objects) and 2 bars (separating 3 varieties)

[last slate]

yea

[noble lance]

that particular example has a mix of all 3 varieties, but you could shift around the bars to show anything right

like (★ ★ ★ ★ ★ ★ ★ | |) would mean all stars are of the first variety

or you could do (| ★ ★ ★ ★ ★ ★ ★ |) for all stars being of the second variety

or really anything

in a sense, and this is the key part, all ways to arrange these stars and bars correspond with all the ways you could pick out objects from the different varieties

[last slate]

so the point is you count how many ways you could have 7 of 3 varieties?

[noble lance]

bingo

so now our job is a different problem, but if we can do this then we can solve the original problem

how many ways can we arrange the stars and bars?

[last slate]

And are there duplicates

[last slate]

ngl i cant seem to think of it like that, i can only see it as “how many ways can you arrange m varieties to make up n “

but thats correct tho right

[noble lance]

well, it's really just stars and bars we're looking at now

a particular configuration of stars and bars corresponds with a particular selection of objects of certain varieties

in the latter, there's no question of "arrangement" right, it's only about selection

but by turning this problem into stars and bars, now it's about arrangement (and no longer about selection)

does that make sense? or if not let me know where you got stuck @last slate

[last slate]

ngl its not

im trying to understand the formula

[noble lance]

forget the formula, we're going to build up to that

[last slate]

because that helps me understand what were counting

[noble lance]

does the correspondence of the two problems make sense

the key point here is that counting the configurations of one problem is the same as counting the configurations of the other problem

[last slate]

its not really clicking for me

the bars and stars thing

[noble lance]

gotcha

let's look at this configuration again: ★ ★ ★ ★ | ★ | ★ ★

[last slate]

okay

[noble lance]

let's just say the objects are balls and the three varieties are red, green, and blue

[last slate]

alright

[tribal heron]

Hmmmm

-# beggars/ stars and bars

[noble lance]

this arrangement of stars and bars (where all stars are the same, all bars are the same, it's just the order you care about) corresponds with 4 red balls, 1 green ball, and 2 blue balls

[last slate]

wait is this the same as mapping to a string of bits?

[noble lance]

in what sense?

[noble lance]

oh huh okay

sure, you could think of the stars as 0s and bars as 1s?

[last slate]

yes i get it now

so like what are we doing with tht stars bars/0s and 1s

[noble lance]

right

so recall that every configuration of stars and bars corresponds with a selection of objects from the available varieties (and vice versa)

[last slate]

yea

[noble lance]

and we want to count the number of selections

so if we count the number of stars and bars arrangements, we will get our answer

[tribal heron]

My sir taught me an excellently simple way of beggars method

[noble lance]

so the question is this: how many ways can we arrange the stars and bars?

[last slate]

yeah how do you know

[noble lance]

yep okay so

notice how the stars and bars are just kinda in a line

[last slate]

yea

[noble lance]

in the case above (★ ★ ★ ★ | ★ | ★ ★) there's 7 stars and 2 bars

can you recall how many ways there are to arrange 9 objects?

[last slate]

are we doing a permuation

[noble lance]

mhm

[last slate]

9!

waya

ways

[noble lance]

amazing

but

there's a problem; we have duplicates

[quiet grove]

btw brodie in pnc u have to be able to think of how to apply the formulae creatively to get the harder questions right
so it would help you if you learnt more indirect ways of applying so you get the feel

[noble lance]

consider
★ ★ ★ ★ | ★ | ★ ★
and
★ ★ ★ ★ | ★ | ★ ★
are these permutations of stars and bars the same or different?

[last slate]

could be different

like could be a duplicate

they are the same tho

[noble lance]

good, and importantly, they correspond to the same selection of objects from varieties

[last slate]

yea

[noble lance]

but actually, I swapped the last two stars and also swapped around the 2 bars

[last slate]

yea

[noble lance]

turns out that didn't make a difference in the end right

so we have to account for this

[last slate]

that too

[noble lance]

it turns out that we have to divide by all the permutations of stars, as well as all the permutations of bars

so let's quickfire this; we have 7 stars and 2 bars

how many ways can we permute 7 objects?

[last slate]

thats 9

but I thought we were doing it all together

[noble lance]

nope, we're doing each type of object because indistinguishability is important here

when you have a permutation of objects where some are indistinguishable, you have to go through and divide by the permutations of only the indistinguishable groups

in this case we have a group of 7 indistinguishable stars, and a group of 2 indistinguishable bars

[noble lance]

(consider if I swapped a star and a bar, then you actually would be able to tell the difference)

with me still? @last slate

[last slate]

yes

im just tryna think of it

[last slate]

wait so

what would be wrong with doing permutations on the set of stars and bars together

like what would that give

[noble lance]

I'll show you a concrete example that's a bit smaller, so you can see what I'm talking about

[last slate]

okay

[noble lance]

suppose I want the permutations of (1, 2, 3, 3), but the 3s are indistinguishable so swapping them gives me the same permutation

[tribal heron]

-# isn't stars and bars just the distribution of alike things?

[quiet grove]

yeah

[last slate]

true so how does doing them separate solve that

[noble lance]

if I just take a factorial, I will get 4! = 24 permutations, and this is small enough to list out
1 2 3 3
1 2 3 3
1 3 2 3
1 3 3 2
1 3 3 2
1 3 2 3
2 1 3 3
2 1 3 3
2 3 1 3
2 3 3 1
2 3 3 1
2 3 1 3
3 1 2 3
3 1 3 2
3 2 1 3
3 2 3 1
3 3 1 2
3 3 2 1
3 1 2 3
3 1 3 2
3 2 1 3
3 2 3 1
3 3 1 2
3 3 2 1

if you look through this list, you will notice duplicates that I did not intend to count

[last slate]

true

[noble lance]

in fact, I have counted exactly double because swapping the 3s gives me exactly 1 extra copy per permutation

so how do I account for this?

[last slate]

you divide by r!

which maps to a single set

for each duplicate

[noble lance]

what's r here?

[last slate]

2!

[noble lance]

ah yep okay that's right

[last slate]

wait

its 4!

[noble lance]

which, to be clear, is the number of permutations of the indistinguishable objects

does that make sense?

[noble lance]

well, if I divide 4! by 4! I end up with 1 permutation overall which doesn't seem right

[last slate]

but r is the number of permutations

Whats the number of permutations here

[noble lance]

I kind of want to ask you that, cuz I'm not super sure where r came from-

[noble lance]

this is what I want to count

[last slate]

I don’t get it

were doing 4 permutations

[noble lance]

mhm

so we have 4 objects to permute

we did 4!, and that gave us 24 permutations

[noble lance]

but when we list out the permutations, we see that there are duplicates

i.e. we are over-counting, and we need to account for this

what should our strategy for that be here

[last slate]

That was like dividing by k

where k is some value

[noble lance]

mhm, and how do we find that value

[last slate]

idk 🤣

I thought i knew

[noble lance]

hahahah

well

notice how if I take one of the permutations

say 3 1 3 2

and I swap the two 3s, I end up with the permutation 3 1 3 2

which, since those are indistinguishable, is actually the same permutation

[last slate]

yes

I agree

[noble lance]

right

suppose a different scenario where I was instead permuting 1 2 2 2

any permution that keeps the 1 there and moves around the 2s will also be the same permutation

[last slate]

right

[noble lance]

because any permutation of 2s leads to the same overall permutation, and I have 3 of them, I have to divide by 3! (since that's how much I'm overcounting by)

[last slate]

yes

[noble lance]

in this case, any permutation of the 3s leads to the same overall permutaion, and I have 2 of them, I have to divide by 2!

so the answer for permutations of (1, 2, 3, 3) in the end is 4!/2! = 12
1 2 3 3
1 3 2 3
1 3 3 2
2 1 3 3
2 3 1 3
2 3 3 1
3 1 2 3
3 1 3 2
3 2 1 3
3 2 3 1
3 3 1 2
3 3 2 1

does this make sense? if not please tell me, because this is crucial before going back to stars and bars

[last slate]

so ⇒ / by variety1! * variety2! * variety3!

?

[noble lance]

exactly!

[last slate]

because this is what i was thinking

[noble lance]

yeah I think this is something different, I'm not really sure what the working out there means

[last slate]

its a k to 1 mapping

for counting subsets

[noble lance]

oh right okay yeah counting subsets is a different problem

okay did everything above make sense? shall we go back to stars and bars

[last slate]

yes

[noble lance]

let's go back to this example: ★ ★ ★ ★ | ★ | ★ ★

we have 9 objects: 7 stars, and 2 bars

how many ways to arrange them?

[last slate]

9!

[quiet grove]

sorry to wander off topic but is the discord.py server from your tag a public server? i cant find it on the explore page

[quiet grove]

remember that 7 stars are alike and 2 bars are alike

[noble lance]

yep should be, just click on it no?

[quiet grove]

maybe it has an 18+ filter? i dont see a go to server button

[noble lance]

yep, so as @quiet grove mentioned, we know that there are some indistinguishable objects here, so we have to divide by the permutations within each variety - what do you get when you do this?

[noble lance]

no idea, the invite is in the library's documentation page - https://discordpy.readthedocs.io/en/latest/

[last slate]

Ok so lemme do it like this

you want how many ways to arrange

000010100

Such that 0s coresponds to different varities and 1s are separaters

So then you wanna know how many ways to rearrange the 0s and 1s so you know that you can arrange the 0s P(9, 7) /7! * P(9,3)/3!

Hope i didnt cook myself there

= 9!/7! * 9!/3!

@noble lance

[noble lance]

what happened to this?

[last slate]

Probably wrong

[last slate]

I forgot what i was thinking

[noble lance]

hahahah

[last slate]

3 hours of sleep side effects

[noble lance]

I mean you're on the right track but I think you've fallen back onto some old miscalculation habits

you shouldn't be doing math on low sleep anyway

[noble lance]

let's go back to this idea, cuz this was spot on

when you're looking for permutations of objects with indistinguishable varieties

you do (total number)! / (variety1! * variety2! * ...)

that's where we got to yeah?

@last slate

[last slate]

yea

[last slate]

yes

[noble lance]

right, so let's just focus on what the numbers should be to plug into this expression

we have ★ ★ ★ ★ | ★ | ★ ★

what's the total number of objects here?

[last slate]

9

[noble lance]

great, so (9)! / (variety1! * variety2! * ...)

how many varieties of objects are there here?

[last slate]

3?

wait

4 1 and 2

[noble lance]

I literally mean like in ★ ★ ★ ★ | ★ | ★ ★

(what is this thing called again)

[last slate]

7?

[scenic star]

Stars and bars?

[quiet grove]

7 of what

[last slate]

Sorry bro can we backup a little

theres something i need to understand

[quiet grove]

which part

[last slate]

the part where we need to rearrange 000010100 certain times

so we need to see how many ways we can rearrange these

right

without duplicates

[quiet grove]

yeah

[noble lance]

mhm

[last slate]

okay so we know we can arrange them 9! Times

[quiet grove]

and then?

[last slate]

now im tryna understand the duplication part

Of why it happens then why does dividing that way gets rid of it

[noble lance]

remember the (1, 2, 3, 3) example?

[last slate]

Yes

so is it like

in each unique position the 1s can be duplicated 7 times

I mean 0s

and every unique position for the 1s we have 2! Extra copies of that

and we dont want those

[quiet grove]

right

[last slate]

so /7!*2!

right

[noble lance]

yep

[quiet grove]

correct

[last slate]

so the whole thing becomes 9!/2!*7!?

[quiet grove]

exactly

W

[last slate]

Yesssisisisisi

[noble lance]

🔥

[last slate]

but wait

is this how you always do it

and like what about the multi set thing

or is that counted for

[quiet grove]

what multiset are you referring to

[noble lance]

for permutations with repeated (indistinguishable) objects, yes

[last slate]

like in general i guess

[last slate]

and that works for the cupcake example

[noble lance]

i mean all these examples are equivalent right

so yeah you can generalize this reasoning in a way

okay so we ended up with 9! / (2! * 7!) in the end right

but we have to work this back to the first problem, and then it will finally all come together

we got 9 as the total number of objects, 2 as the number of bars and 7 as the number of stars yeah?

so in general, we can write something like (stars + bars)! / (stars! * bars!)

makes sense? @last slate

[last slate]

My bad phone died

[last slate]

yea

[last slate]

yes

i see what’s happening

and then we get all slots for how we can get all possible unique arrangements corresponding to each star or 0

[noble lance]

yep

so now if we recall how the stars and bars correspond to the original question

[noble lance]

if we go back to this message, we could work out that stars = n and bars = m - 1, yeah?

[last slate]

yes

btw

how does this also apply to the math problem

like the x1+ x2 +x3 +x4 +x5 = 20

for example

[noble lance]

sorry what do you mean by this?

[noble lance]

plugging this in, we get that the number of ways to select n objects of m varieties is (n + m - 1)! / (n! * (m - 1)!)

[noble lance]

um, this is a different problem?

[last slate]

same concept tho no?

im asking how to apply the stars and bars thing on it

[noble lance]

I'm looking at this problem you sent earlier

[last slate]

yes

its a different problem

[noble lance]

ah sure okay

well

try to think about what a "star" represents here and what a "bar" represents here

[last slate]

Stars or 0s represent positions

bars or 1s represent seperators

[quiet grove]

first you need to understand why we can use stars and bars, so try to think of what "undistinguishable object" are we distributing

[last slate]

i don’t understand what they are tbh

[quiet grove]

whats your country's currency?

[last slate]

dollars

[quiet grove]

okay

so imagine 30 as 30 dollars

[last slate]

Okay

[quiet grove]

you can write 30 dollars as 30 1 dollar notes/coins

correct?

[last slate]

yes

[quiet grove]

would you know the difference between 2 coins?

[last slate]

no

[quiet grove]

right

so the coins here are the indistinguishable object that we are trying to divide into 5 people

so basically x1+x2+x3+x4+x5=30 is dividing 30 indistinguishable coins between 5 people

so if you use the stars and bars formula that @noble lance helped you understand, you would get the right answer

did you get it?

[last slate]

so i have 5 varieties

and im tryna see how many ways i can make 30

[quiet grove]

yeah

by giving the 5 varieties different amounts

exactly

so you got it?

[last slate]

thats 30 slots

[last slate]

id need to know my solutions tho?

[quiet grove]

wdym

[quiet grove]

5 slots 30 coins

[last slate]

lenme get some rest and get back to it 🤣

@quiet grove @noble lance i appreciate you guys

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

update

[last slate]

need help on this one

the want at least one kind of each variety in every combination

[midnight plank]

@last slate Has your question been resolved?

[last slate]

<@&286206848099549185>

yoooo

[hybrid crane]

You can initially pick 7, all of different varieties. The order doesn't matter so there's only one way to choose that 7.

That leaves 5 donuts to pick, still from 7 choices (with replacement of course)

[lilac solstice]

List out the combinations.

[lilac solstice]

Then do calculations.

[last slate]

I don’t understand the part about havin at least 1 selection

btw im doing bars and stars method

[last slate]

@lilac solstice

[hybrid crane]

Since the order doesn't matter, I could fulfill the requirement of having 1 of each kind by just picking 7 donuts of different variety

Then we are left with 5 more donuts to pick

That's where you would use stars and bars

[last slate]

so the thing is im trying to think of this via rearranging like 111111000000000000

same to stars and bars but 0s are strs

1 are bars

how do i do that now

[last slate]

what’s gonna be different in my rearranging tho i dont get it

🤔

[hybrid crane]

In my mind, you would only forget the first 7 (it'll just add a factor of 1 to your answer). We want to choose 5 items from 7 with repetition/replacement.

As for the layout, there will always be 5 bars and 6 stars - so you can figure it out from there

Had to refresh on stars and bars sors

I think you could do it in full but it'd end up the being the same computation anyway

[lilac solstice]

Actually I would just list down the possible combinations (not all but the available spaces as empty lines) then calculate.

[last slate]

so youre looking at 5 donuts?

[hybrid crane]

At picking 5 from 7, yes

(With the aforementioned considerations)

[coarse wing]

gg leon you finally improved

[last slate]

i dont get it at all

[last slate]

oh yeah thank God

thank u

but im still struggling with restrictions

<@&286206848099549185>

Can someone help me understand the restrictions im still confused

i mean why are we shortening the set

[midnight plank]

@last slate Has your question been resolved?

[last slate]

<@&286206848099549185> yall recommended i just use this table?

[latent wadi]

Yes but i recommend that you should understand where these formulae come from

[midnight plank]

@last slate Has your question been resolved?

[sudden bronze]

yo can yall pray for me i got 3 exams tmmr i pray the owner of doge but better does well in his exams

[midnight plank]

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[midnight plank]

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[vale charm]

question 2

[wooden badger]

What hv u done?

[vale charm]

its abit messy

so i did it to try solve quesiton 2

<@&286206848099549185> sorrry kinda tight of time🥲

[wooden badger]

Uhh

Theres a triangle

[vale charm]

?

for question 2?

[wooden badger]

ABC

You can find the angle B using angle sum property

[vale charm]

oooooh

THANKSKSS

[wooden badger]

Yassss

[vale charm]

.close

[midnight plank]

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[vast summit]

Can someone please check my problem in #forum ??