help-49

[neon bone]

source is wiki but I will try to explain

For the Moment Generating Function ($\mathbb{E}[e^{tX}]$) to exist, the tails of the probability distribution must decay to zero at least as fast as an exponential function (like the Normal distribution) however 'heavy-tailed' distributions like the Cauchy distribution decay only at a polynomial rate (like $1/x^2$). Exponential growth ($e^{tx}$) always overpowers polynomial decay. Therefore, for all cases where $t \neq 0$, the integral (expected value) diverges to infinity

[grand pond]

ShaddowDagger

[neon bone]

That's the most popular but u can find another distributions like that with some research that may take some time

[zealous schooner]

ok I didn't read yours but I think I did in fact come up with a discrete example

(I want a discrete example since that's what we've covered so far, even though I know how continuous distributions work)

We can make a random variable $X$ such that [ \mathbb P(X=n) = \mathbb P(X=-n)=\frac{3}{\pi^2}\frac1{n^2} \qquad \text{for }n\in\bN\setminus {0} ]

[grand pond]

kheer257

[neon bone]

yep thats totally works

[zealous schooner]

If $t>0$ then [ \mathbb E\left[e^{tX}\right] = \frac{3}{\pi^2} \sum_{n\in\bN\setminus{0}} \frac{e^{nt}+e^{-nt}}{n^2} \ge \sum_{n=1}^\infty \frac{e^{nt}}{n^2} \to\infty ]

[grand pond]

kheer257

[zealous schooner]

and we just take the other term for t < 0

that series diverges because of the root test

[neon bone]

thats right

and also I agree with you not reading it

Sometimes reading things like this can get confusing

[zealous schooner]

my apologies lmao

[zealous schooner]

hmm so we want the density to decay faster than all exponential decay but slower than polynomial

[neon bone]

Actually the example you wrote fits this very well

[zealous schooner]

we can just make it continuous somehow

well we can take something like e^(-sqrt(x))

yeah just take a normalised version of e^(-sqrt(|x|))

that converges right

,w integral over (-infty, infty) of e^(-sqrt(|x|))

[grand pond]

Results provided by WolframAlpha

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Requested by kheer257

[zealous schooner]

yeah

[zealous schooner]

Yeah so [ \mathbb E\left[e^{tX}\right] = \frac14\int_{-\infty}^\infty e^{tx-\sqrt{|x|}} \dd{x} \to +\infty ] for all $t\ne 0$

[grand pond]

kheer257

[zealous schooner]

neat

[neon bone]

You've actually summarized the whole issue. We can write as many statements as we want in this style

[neon bone]

In this

[zealous schooner]

do you mean as many functions?

[neon bone]

yep

[zealous schooner]

we could also have something like e^(1/x)?

no that totally doesn't work lmao

[neon bone]

Now the question that is bothering me is whether we have to host the radical to balance the speed of the expression exponentially

Maybe I need to get help too

If you make the exponent positive ($e^{...}$), the function will either explode or remain constant, resulting in no distribution. You should make the exponent negative, but lower the degree (like $e^{-x^{0.5}}$, $e^{-x^{0.1}}$).

Do u think this sentence correct?

[grand pond]

ShaddowDagger

[zealous schooner]

yeah

I think it should be

like $\int_{-\infty}^\infty e^{-\abs{x}^t} \dd{x}$ converges for all $t>0$

[grand pond]

kheer257

[zealous schooner]

and $\int_{-\infty}^\infty e^{tx-\abs{x}^r} \dd{x}$ does not converge for any $t\ne 0$ if we choose $r\in (0,1)$

[grand pond]

kheer257

[zealous schooner]

because at either +inf or -inf the dominating term is just e^(tx) which diverges

[neon bone]

At this stage your knowledge began to defeat me

but yeah

[zealous schooner]

haha I just like integrals

Are we sure this is true for all t? I feel like it's only true for t>0

[neon bone]

yeah def. for t>0

[zealous schooner]

hey frosst

[subtle blaze]

$\int_x^\infty f_X,dx \leq \int_\bR e^{ty}e^{-tx}f_X, dy$

Idk just writing it out

[zealous schooner]

f_X on the right asw

f_X(y) that is

[grand pond]

frosst

[subtle blaze]

Ye forgor

[zealous schooner]

yeah so if we see the difference it's [ M_X(t)e^{-xt} - \mathbb P(X>x) = \int_{-\infty}^x e^{(y-x)t}f_X(y) \dd{y} + \int_x^\infty \left(e^{(y-x)t}-1\right) f_X(y) \dd{y} ]

If $t>0$ then obviously both terms are positive

[subtle blaze]

There you go

[grand pond]

kheer257

[zealous schooner]

what happens when t<0

[neon bone]

probably just changes direction

[zealous schooner]

maybe

I think we can assume x=0

[subtle blaze]

Well does it work if you put x-y

Instead of y-x

Yes right cos everything are still nonnegative functions

Just slip the minus sign inside

[zealous schooner]

but then the expectations are all whack

[subtle blaze]

Oh idk about the -1

[zealous schooner]

M_x(t) just becomes M_x(-t) which doesn't help

[subtle blaze]

Idk if e^(x-y)t -1 is positive

[zealous schooner]

it's negative

because y>x

oh well if t < 0 then it's positive

[subtle blaze]

Wait surely that’s fine

M_X(t) = E[e^tX] if you put negative it’s the same as reversing X

Like doing -X

If $X\in \mathcal L^1$ then so is $-X$

[grand pond]

frosst

[zealous schooner]

what's L1

[subtle blaze]

Integrable functions

[zealous schooner]

bruh

[subtle blaze]

$X\in\mathcal L^p \iff \int_\Omega |X|^p,d\mu <\infty$

[grand pond]

frosst

[zealous schooner]

I have in fact seen this before

For Hölder's inequality

[subtle blaze]

For real random variables $\int_\bR |X|,dx < \infty$

[grand pond]

frosst

[zealous schooner]

So you're saying if $t<0$ we consider $-X$ and $-x$ because $M_X(t)=M_{-X}(-t)$

[grand pond]

kheer257

[subtle blaze]

Ok that’s a poor abuse of notation

[zealous schooner]

but then P(X > x) = P(-X < -x)

[subtle blaze]

Yeah M_X(-t) = M_-X(t)

Well I’m just saying if one is finite the other is also finite

[subtle blaze]

For this at least

Maybe I’m going off on the wrong tangent now that I look at it

[zealous schooner]

And also f_{-X}(y) = f_X(-y)

[zealous schooner]

honestly now I think it might work

no I think you just flip all the integrals

the problem remains

low-key I think this is only true for positive t

I might just be getting trolled by my tutor

[subtle blaze]

Well find a counterexample I guess

[zealous schooner]

yeah I just took X to be an exponential distribution

It isn't true for x < 0 if t < 0

my tutor is such a troll

actually I think the inequality just reverses if t < 0

but i can't be bothered to prove it

[subtle blaze]

Isn’t exponential only defined for positive x

Although I guess it can be zero for negatives

[zealous schooner]

yeah

I mean P(X>x) is just 1 if x < 0

[subtle blaze]

Yeah goodness me I haven’t done probability in like 1.5 years

[zealous schooner]

it's all just integrals

[zealous schooner]

I mean the inequality isn't really useful for negative t anyway

the next part is to minimise the right side for different distributions to get optimal tail bounds

and surely that will be for t > 0

[neon bone]

It's so much fun to look at the conversation and search for the most effective painkiller for my brain

[neon bone]

something I can agree

[zealous schooner]

alright I think it's been too much maths for one day

thanks for y'all's help

.close

[midnight plank]

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[midnight plank]

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[grand pond]

Renato

[latent wadi]

What have you tried

[tidal turret]

@latent wadi

[tidal turret]

you here?

i got stuck near the end

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[pine wave]

Yeah so here you just need to combine

[tidal turret]

care to elaborate

[pine wave]

You see it's either 2^0 or 2^1, you see 5^4

So you can have 5^4 or 2*5^4

[tidal turret]

how does that work

can we start from scratch

[pine wave]

case on largest power of 2 that divides d

And from there calculate d

[tidal turret]

how

[pine wave]

Your second slide says that it's either 2^0 or 2^1 right

Largest power of 2 that divides d

[tidal turret]

it can't be 2^1 because then a = b (mod 2) but gcd(a,b) = 5

correct so far?

what i mean is, they both can't contain a factor of 2 simultaneously in the second slide

and they are congruent under mod 2
both a = b (mod 2) so both need to be odd

because if one is even the other one aswell and then gcd(a,b) is not 5

@pine wave Hopefully u understand what i mean

[tidal turret]

what i mean is, in second slide what we are doing is checking in which case does d have a factor of 2, and that's if and only if a is congruent to b under mod 2

then we use that gcd(a,b)=5, so both can't be a 0 under mod 2 while at the same time being congruent to each other because then gcd(a,b) would be 10 instead of 5

[pine wave]

Yeah so there are two possibilities, both a b odd giving a factor of 2, otherwise no factor of 2

[tidal turret]

yesss you got it

that's right

[tidal turret]

this shit is hard to summarize in one sentence, no?

[pine wave]

so basically you have two cases

Has factor of 2 or no factor of 2

[tidal turret]

yes

[pine wave]

And then you try to solve for a, b in each case

As in, find one explicit example

[tidal turret]

wait but we still need to examine the 5^k

when i look at the expressions under mod 5^4 i found that either k or q need to contain a 5, but not both at the same time because otherwise gcd(a,b) = 5^2

that's the third slide

i find that 5^4 | a^4 + b^4 for any a,b in Z because gcd(a,b)=5

however when does 5^4 | 2ab^2 ?

well since we know a = 5k and b = 5q, with gcd(k,q) = 1

5^4 | 2ab^2 <=> 2ab^2 = 0 (mod 5^4) <=> ab^2 = 0 (mod 5^4) <=> 5^3 × k × q^2 = 0 (mod 5^4) <=> kq^2 = 0 (mod 5)

so since gcd(k,q) = 1, this is only possible if one has a 5 and the other doesn't @pine wave

this is start the third slide is saying

so basically, 2 divides d <=> a = 1 (mod 2) and b = 1 (mod 2)

[pine wave]

yeah you eventually find 5^4 is the only power of 5 you can have dividing d wait mistake

[tidal turret]

one step at a time, there is no rush

[pine wave]

Oh wait oof

yeah it should be only 5^3 or 5^4

And then you need to construct the 4 cases

[tidal turret]

yes im still trying to summarize what we have

2 divides d <=> a = 1 (mod 2) and b = 1 (mod 2)
2 does not divide d <=> ?

[pine wave]

a even b odd or a odd b even

It's basically the logical negation of your found case

[tidal turret]

well the actual negation would be, a even or b even, but it's an exclusive or, because gcd(a,b)=5

[tidal turret]

yes

[tidal turret]

either a even or b even but not both

2 divides d <=> a = odd and b = odd
2 does not divide d <=> (a = even and b = odd) or (a = odd and b = even)

[tidal turret]

should i look for divisibility with 5^3 first?

because i straight up jumped to divisibility with 5^4

like, i should have started with divisibility by 5, 5^2, 5^3, 5^4, etc

[tidal turret]

5^3 always divides a^4 + b^4

[pine wave]

Yeah

[tidal turret]

5^3 always divides 2ab^2

irregardless of a,b FORALL AB IN Z

[pine wave]

*regardless

or irrespective

[tidal turret]

2 divides d <=> a = odd and b = odd
2 does not divide d <=> (a = even and b = odd) or (a = odd and b = even)
5,5^2,5^3 divides d <=> ALWAYS

[pine wave]

So now you need an example of
d=2^0 5^3
d=2^1 5^3
d=2^0 5^4
d=2^1 5^4

[tidal turret]

holdon, i need to check the conditions on a,b, when 5^4 | d

same with 5^5, and etc, until i get something like 5^k never divides d

[tidal turret]

just, patience, we are getting there

2 divides d <=> a = odd and b = odd
2 does not divide d <=> (a = even and b = odd) or (a = odd and b = even)
5,5^2,5^3 divides d <=> ALWAYS
5^4 divides d <=> (5 divides k and 5 does not divide q) or (5 divides q and 5 does not divide k)
5^4 does not divide d <=> (5 does not divide k) and (5 does not divide q)

@pine wave help with negation

[tidal turret]

5^5 is giving me trouble

ah, for examining 5^5 | d, we need to take as granted 5^4 | d

so one of the conditions for 5^5 | d to hold is that,
5 | k xor 5 | q

for 5^5 | d we need to assume either k = 0 (mod 5) or q = 0 (mod 5) but not both

so check 5^5 | 2ab^2 we get

[pine wave]

hmm 5^5 shouldnt happen right

[tidal turret]

HOW

[pine wave]

cases: 5^2 div a but not b fail/ b but not a, fail
Neither, then a^4 and b^4 are 5^4 mod 5^5

[tidal turret]

what?

[pine wave]

You can check that a^4+b^4 is 625 or 1250 mod 3125

might be easier to divide by 625 first

[pine wave]

yeah 5^5 shouldnt happen

[tidal turret]

if it happens then k is not cop q

what i mean is that if it happens then gcd(a,b) is not 5

[pine wave]

okay doesnt really matter which contradiction you get

[tidal turret]

bec gcd(a,b ) = gcd(5k,5q) = 5.gcd(k,q) = 5

[tidal turret]

the idea is that 5,5^2,5^3,5^4 holds for some conditions, but 5^5 never and by consequence 5^k for k > 4 never

[tidal turret]

2 divides d <=> a = odd and b = odd
2 does not divide d <=> (a = even and b = odd) or (a = odd and b = even)
5,5^2,5^3 divides d <=> ALWAYS
5^4 divides d <=> (5 divides k and 5 does not divide q) or (5 divides q and 5 does not divide k)
5^4 does not divide d <=> (5 does not divide k) and (5 does not divide q)
5^5 divides d <=> NEVER

[tidal turret]

gcd(a,b) = c => c | a and c | b

[pine wave]

So now combine the conditions to get examples

[tidal turret]

this is the hard part

[pine wave]

How do you get 2 divides d and 5^4 divides d?

[tidal turret]

both simul you mean?

5^4 | d <=> 5 | k xor 5 | q
2 | d <=> a = odd and b = odd
2.5^4 | d <=> (5 | k xor 5 | q) and ( a is odd ) and (b is odd) and gcd(a,b) = 5

@pine wave

[pine wave]

yeah now get an example of a and b that satisfy these conditions

[tidal turret]

(a,b)=(3 × 5, 5) = (15,5) => 2.5^4 | d

[pine wave]

b cant be 7

[tidal turret]

yes, 5 | a and 5 | b, because gcd(a,b) = 5

my bad xD

[tidal turret]

,w gcd(a^4+b^4,2ab^2), when a = 15, b = 5

[pine wave]

You got 2*5^3

[tidal turret]

what happened

[pine wave]

Check the requirements for 5^4

You need 25|a xor 25|b

[tidal turret]

how?

[pine wave]

5|k means 25|a right

[tidal turret]

but

2 divides d <=> a = odd and b = odd
2 does not divide d <=> a = even xor b = even
5,5^2,5^3 divides d <=> ALWAYS
5^4 divides d <=> 25 | a xor 25 | b
5^4 does not divide d <=> (25 does not divide a) and (25 does not divide b)
5^5 divides d <=> NEVER

[tidal turret]

how is the second one stronger?

[pine wave]

a=5k right

[tidal turret]

ye

[pine wave]

you have
5^4 divides d <=> 25 | a xor 25 | b
which (a, b)=(15, 5) doesnt follow this condition

[tidal turret]

ab = 25,5

[pine wave]

Yeah try it out

[tidal turret]

,w gcd(a^4+b^4,2ab^2) = 2 * 5^4, when a = 25, b = 5

[pine wave]

Yeah so now you obtained 2*5^4

Next you have to obtain 5^3 and 5^4

[grand pond]

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Requested by renatoxbz

[tidal turret]

what are the possibilities?

[pine wave]

You need to combine "2 doesnt divide d" with "5^4 divides d"

[tidal turret]

wait

A = {2,5,5^2,5^3,5^4}
d = AxA \ {(2,2), (5^2, 5^2), (5^3, 5^3), (5^4,5^4)}?

@pine wave

[pine wave]

wait why

(Check that you cant have d=5^2)

[tidal turret]

i am trying to find all the possible values of d

[tidal turret]

wdym?

[pine wave]

the underlying idea is you choose which power of 2 to divide d (2^0 or 2^1) and which power of 5 to divide d (5^3 or 5^4)

[tidal turret]

so how do i count how many

2 choices for each 5^k with k in [0,4]

[pine wave]

2 choices for power of 2: 0 1
2 choices for power of 5: 3 4
2*2=4 choices in total

[tidal turret]

why do you say this

[pine wave]

think of this as choosing the largest power dividing d

[tidal turret]

how do you know 5^2 does not divide d

that seems out of the blue

[pine wave]

5^2 always divides d, but 5^2 isnt the largest power of 5 that divides d

[tidal turret]

what

[tidal turret]

what do u mean

2 divides d <=> a = odd and b = odd
2 does not divide d <=> a = even xor b = even
5,5^2,5^3 divides d <=> ALWAYS
5^4 divides d <=> 25 | a xor 25 | b
5^4 does not divide d <=> (25 does not divide a) and (25 does not divide b)
5^5 divides d <=> NEVER

[tidal turret]

the idea with gcd is that we are looking for the gcd, the greatest that divides them both

[tidal turret]

I see

[tidal turret]

dude this exercise is like constructing gcd out of sticks and stones correct?

hardest shit I have done in a while

[pine wave]

maybe just needs more practice

[tidal turret]

you think so? it gets better?

[tidal turret]

basically either
d | 2
d | 5^3
d | 2 × 5^3
d | 2 × 5^4
d | 5^4
?

no, d | 2 is impossible because we always divide 5^3 no matter what

wait, so it is

d1 = 5^3
d2 = 2 × 5^3
d3 = 2 × 5^4
d4 = 5^4
?

@pine wave

that's all the possibilities

[pine wave]

Yeah

[tidal turret]

ok great

d1 = 5^3 <=> a = even xor b = even

(a,b) = (10, 15)

[pine wave]

Yeah

[tidal turret]

,w gcd(a^4+b^4,2ab^2) = 5^3, when a = 10, b = 15

d2 = 2 × 5^3 <=> a is odd and b is odd and gcd(a,b) = 5 and gcd(k,q) = 1

(a,b) = (5,15)

,w gcd(a^4+b^4,2ab^2) = 2*5^3, when a = 5, b = 15

d3 = 2 × 5^4 <=> (a = odd) and (b = odd) and (25 | a xor 25 | b)

(a,b) = (5,25)

,w gcd(a^4+b^4,2ab^2) = 2*5^4, when a = 5, b = 25

d4 = 5^4 <=> (a = even xor b = even) and (25 | a xor 25 | b)

(a,b) = (10, 25)

,w gcd(a^4+b^4,2ab^2) = 5^4, when a = 10, b = 25

[grand pond]

Results provided by WolframAlpha

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[tidal turret]

@pine wave

sir I appreciate the help... I cant thank u enough

this kind of exercise always gives me trouble

I have studied a bunch but it still gives me trouble, sometimes you just need more time to make your head get it you know?

[pine wave]

alr

[tidal turret]

like, I give it a ton of effort but sometimes shit just doesnt click and I get stuck

this kind of exercises always gives me a brain fart

[tidal turret]

if it weren't for u I would have give up already with this, like is so hard 😪

I need to practice more of this exercises because they are not that impossible but is easy to get stuck

also, midway you need to know a lot of properties and whatnot

[tidal turret]

also, this part of providing and example for each d is incredibly hard

you need to characterize d and look for all the modulos

hardest shit I have ever done in my life

like, even if you find d | 2 * 5^4

you can still get stuck finding the values of d

maybe im just bad at number theory

but that's how I feel it is

you need to internalize everything

[tidal turret]

alr

.solved

[midnight plank]

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[stoic smelt]

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples

[stoic smelt]

What did I do wrong

[runic hamlet]

7x and 14x are not two consecutive multiples

eg for x=2 they are 14 and 28

while you would want 14 and 21

[hexed barn]

hint: use addition to create the second consecutive multiple of 7

[runic hamlet]

well and then later you divided the lhs by 5 and not the rhs so thats another mistake

[stoic smelt]

OOOHHHHH

So x + (x+7)

[hexed barn]

don't forget your 7

[stoic smelt]

So 7x and 7x+7

[hexed barn]

+2?

[stoic smelt]

Yea i aint getting a good grade in maths

[hexed barn]

there we go

[stoic smelt]

Thanks g

Ill be back soon w something else

.close

[midnight plank]

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[midnight plank]

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[stoic smelt]

Same question

[stoic smelt]

What did I do wrong

The answer is not giving a whole no.

[wispy mortar]

+14x?

[lusty python]

It's correct

[lusty python]

Wdym?

Show your full question

[wispy mortar]

huh

[lusty python]

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[wispy mortar]

no its not

[peak marlin]

you went wrong in the second line

[wispy mortar]

49*2 bro what

[lusty python]

I take that back

My bad

[peak marlin]

it should be 49n ^ 2 + 49n ^ 2 + 98n + 49 = 637 in the second line

[lusty python]

But why do you even need a whole number for your result?

[lusty python]

I missed that +7 on the back

Probably my idiotic soul took over my body

[peak marlin]

Because in this case you were asked to find two consecutive multiples of 7. Multiples of 7 are whole numbers that are evenly divisible by 7 (e.g., 7, 28, etc.). So when you solve this problem, you should always end up with whole numbers.

[lusty python]

Who said it needs to be a whole number? It can be an integer!

[wispy mortar]

bro

[lusty python]

???

[peak marlin]

Yeah but then this person with the question got the answer as a root so it can't be right, it can be an integer too, just not irrational 😭

[wooden badger]

its not 1414x

the multiplemultiples are 14,221

21

x would be 2,3

[lusty python]

Time to use this

[lusty python]

!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]

NO ANSWERS

[wooden badger]

oh dang

i didn't know tht okay

[midnight plank]

@stoic smelt Has your question been resolved?

[stoic smelt]

This is the original question

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[stoic smelt]

Oh yeah, I forgot a = 7x not x 😭

[midnight plank]

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[turbid haven]

hi

[turbid haven]

,tex $f_1(t) = 3e^t + e^{t^2}\ f_2(t)=7e^t+e^{t^2}\ f_3(t) = 5e^t + e^{-t^3} + e^{t^2}\n$

[grand pond]

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

[turbid haven]

are solutions to

,tex $y''(t) + p(t)y'(t) + q(t)y(t) = g(t)$

[grand pond]

Joe

[turbid haven]

if y(0) = 1 and y'(0) = 2

solve the ivp

i think we should say like
y1 = f1-f2
y2 = f1-f3
but idk the rest

[visual tiger]

so y1 and y2 are independent solutions to the homogeneous equation

being a linear ODE of order 2

so it makes a basis of the solution space

So, for example, what can you say about y - f1?

[turbid haven]

so the y(t) = u1.y1 + u2.y2?

[turbid haven]

i dont know 😭

[visual tiger]

you didn't follow properly

[visual tiger]

y1 and y2 are not solutions of this

[turbid haven]

oh solutions to this:
y'' + py' + qy = 0

[visual tiger]

exactly

so, how do you get solutions to y" + py' + qy = g, given y1 and y2?

[turbid haven]

is it y(t) = c1y1 + c2y2 + yp in this case?

[visual tiger]

exactly

so just pick some yp

[turbid haven]

yp = u1y1 + u2y2?

[visual tiger]

yp is any solution of THIS problem

any that you know of

[turbid haven]

like f1, f2 and f3?

but since p and q are not constants we should use variation of parameters?

[turbid haven]

and using variation of parametes each should be multipled by some function?

[midnight plank]

@turbid haven Has your question been resolved?

Channel closed

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[midnight plank]

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[last slate]

i need help understanding permutations with repetition

[lyric charm]

do you understand permutations if there's no repeated letters

[last slate]

yes

i do

[lyric charm]

ok cool

let's start with HAPPY bc that one is a bit more straightforward

[last slate]

alriight

[lyric charm]

if you were to pretend the P's are different, like $HAP_1P_2Y$

[grand pond]

Ann

[lyric charm]

then that's 5! permutations

[last slate]

Yes

[lyric charm]

but we have overcounted by a factor of 2!

bc like.. $HAP_1P_2Y$ is not actually distinct from $HAP_2P_1Y$

[grand pond]

Ann

[last slate]

i see

yes

and then?

[wooden badger]

so we divide by 2!

[lyric charm]

this happens with every permutation

every 2! permutations collapse into one

[lyric charm]

thus, this

[last slate]

is that k to 1 mapping

or

[lyric charm]

i mean, if you insist

[last slate]

but i dont get how you put it together

for example we were working with the Ps

Which means we are strictly looking at those

Right

[golden ore]

Hey man, see.
It's pretty easy - here's a problem from me. Try the permutations for HONOLULU.

[last slate]

if they were distinct thats 8!

but they arent

[golden ore]

The trick is just to calculate the total no. of permutations and divide it by no. of repetitions. For example - if a letter is repeating x times, then you just divide the whole factorial by x!

[last slate]

yes but im trying to understand the intuition

could you please check out my intuition and see if im right

im gonna try

to explain it

[golden ore]

Yep, go ahead.

Here's mine - think of it like shuffling the cards. You first think all of them are distinct and you find the permuatations, which is n!, but alas! some repeat themselves and they can rotate among themselves. What's the permutations for x things in the deck shuffling among themselves? - it's x!, right? So to fix the overcount, you just divide the whole by no. of repetitions.

[wise flicker]

Look this, if we treat identical letters as distinct

in HAPPY arrange as if they disticnt =>5!
label first P as P1 and second as P2

One arrangement :
P1HYAP2
P2HYAP1

these are different if P1 and P2 were distinct but, they are identical hence we got repeated cases

[last slate]

but

the repetitions arent equal

[golden ore]

Exactly, I never told you they were.

[last slate]

i know

but idk how to imagine it

[golden ore]

Take HONOLULU, you divide the whole factorial by the factorial of repetitions of O, L and U. Why, you ask? Because they are not distinct characters and they can repeat places with themselves the no. of times they repeat - for example, O can repeat 2 times, and so can L and U.

[last slate]

so 5!

?

i mean 6

[wooden badger]

7!

[golden ore]

Yes, that's correct.

[wooden badger]

8! ÷ 2!2!2!

[golden ore]

Mate, the expression is $8!/(2!\times2!\times2!)$ the three 2! are for O, U and L.

[grand pond]

bavajithu

[wooden badger]

Ding dong

[last slate]

im tryna understand this

[wooden badger]

Those are selections?

[last slate]

ye?

[golden ore]

You're only studying Permutations or you're doing combinations too?

I guess they're taught as a single unit, no?

[visual tiger]

this is another way to get the number of permutations of a word with repetition

[wooden badger]

Mostly done tgt

[visual tiger]

You have 11 placeholders

on which you will place letters

[golden ore]

Basically, permutation is arrangement process and combinations is selection process.

[last slate]

hi

anyone from india?

[quiet grove]

isnt it opposite

[golden ore]

Yeah, yeah, my bad.

[wooden badger]

Yas

[visual tiger]

For example, there are 2 Ps. So there are 11C2 choices for the placeholders that will get Ps

there are now 9 empty place holders left

[last slate]

anyone 10th grader bbse

[visual tiger]

Then there are 4 Is. Thus 9C4 ways to choose placeholders for the Is

[last slate]

i did combinations but not like this

[golden ore]

Are you talking about the notation of combinatorics?

[wooden badger]

This might be a msg suited for #discussion

[visual tiger]

This is a helpers channel. For discussions like these, you can go to #discussion

[last slate]

im talking about scenarios

but yeah

[last slate]

ohk sorry

[last slate]

okk

can someone please help

[golden ore]

I mean, they mentioned in the question "scramble", so it basically means selection + arrangement too, no?

[wooden badger]

Raph is explaining p good or no?

[last slate]

idk man i assume so

[wooden badger]

Scramble is arrangement

[golden ore]

To arrange you first have to select, wrong?

[wooden badger]

Well yeah but its 11 C 11 which is basically one

[golden ore]

Ah, yes. But if you're approaching the problem using the empty spaces method, you'd have to use both.

The method where you assume 11 empty spaces, and then go on filling.

[wooden badger]

The formula is total objects ÷ repetitions of objects

[golden ore]

We all know that. Leon wants to understand the intuition behind it.

[wooden badger]

The logic?

[golden ore]

Exactly.

3 people have tried so far, including myself.

[wooden badger]

Ohh

[last slate]

the way done here specifically

i know this is mapping

[wooden badger]

Imagine layering a sandwich

[gusty falcon]

what do you not get?

[wooden badger]

This is legit the best way to word this method

[golden ore]

Oh my, this method is really good. It's way better than memorizing the formula.

[last slate]

the 11, 9 5 and 1

[gusty falcon]

you have 11 slots, and you want to delegate slots for each letter(s)

[golden ore]

Okay, okay. So see, after you place the two Ps, you have 9/11 places left, right?

[gusty falcon]

pick any letter to first lay out

could be P, I, S, or M

[last slate]

isnt that 11! Ways to arrange P

[gusty falcon]

no?

[wooden badger]

Really?

[golden ore]

So in the 9/11 places left, you place Ss, after which x/11 spaces are left, and so on.

[wooden badger]

P the letter?

[gusty falcon]

imagine you have a 2-set like {X,Y} where X and Y are distinct and 0 < X, Y <= 11 denotes the position of the P's

There's certainly not 11! different {X,Y}'s

[golden ore]

Yes! I was having trouble with this concept earlier, but now it's good.

[wooden badger]

Tht methods more logical but the formula is faster

[golden ore]

The exam I'm preparing for doesn't ask me questions where formulae can be applied directly, so...

[last slate]

okay P

[wooden badger]

JEE?

[golden ore]

Ah, yes.

[last slate]

Thats 2!

right

not 11! That was stupid

[wooden badger]

I stand corrected

[gusty falcon]

there's one way to arrange PP

[wooden badger]

Yayy

[gusty falcon]

you're also answering the wrong question

the 11C2 is a different framing

from just "arrangements"

it's more along the lines of

"how many different 2 slots can you delegate to the 2 P's from 11 possible slots"

[last slate]

We are choosing how many times the repeated letters can br arranged without duplicating

[gusty falcon]

it's not "how many ways to arrange 2 P's"

as you can tell

[golden ore]

The name's famous everywhere.

[last slate]

oh my God this cant be this complicated

its not clicking for me bruh

[golden ore]

It isn't.

[last slate]

I can’t

[gusty falcon]

okay here
{X,Y} where X and Y denote the position of P

X and Y is then any integer between 1 and 11, right?

do you get this part?

[golden ore]

Don't waste any more time trying to "understand". Just grind through problems now.

[wooden badger]

Cmon you have 11 seats . There r 2 Ps . How many seats can they sit in?

[gusty falcon]

like {1,3} means one of the P's goes in the first slot

and one of the P goes in the third slot

[last slate]

11

[golden ore]

Guys, math intuition isn't built like that. Let him grind some problem sets.

[last slate]

they can sit in all 11

[wooden badger]

Suppose the Ps are ppl. 2 ppl 11 seats

[wooden badger]

The exam?

[golden ore]

yeah.

[hallow basin]

That's why JEE is bad for your mental health hahahs

[wooden badger]

My family doesn't even know what tht is

[gusty falcon]

if you can answer "how many different groups of 2's can you make with 11 people" then that's the same as answering "how many ways you can laythe two P's down in 11 different slots"

the answer to the first is standard

[golden ore]

I enjoy every single part of it's preparation except for Chemistry.

[last slate]

2

2 seats

[gusty falcon]

and should lead you to 11C2

[wooden badger]

Exactly

[gusty falcon]

if it doesn't then

[wooden badger]

How would u choose those 2 seats?

[hallow basin]

Understanding is key, I barely grinding , like an hour per day on average and still have top grade lol

[golden ore]

Chemistry never sticks for some reason, especially Organic Chemistry.

[wooden badger]

Tht cant be turu

[last slate]

i know they can sit in 10 different ways right

[wooden badger]

11C2

[hallow basin]

-# that's true, way back like in 2 months ago, I did math hw like 15min/day and I spent the rest playing MC 🥀

[wooden badger]

That's selection have u been taught th?

[last slate]

I think?

[wooden badger]

Which gradeTT

[last slate]

i know permutation and combination

and selecting a ⊂

subset

[wooden badger]

Combination is selection

Yesyes

So do u understand where the 11c2 came frm?

[golden ore]

You're in uni, I suppose?

[gusty falcon]

if you knew how to "select subsets" then why is there confusion

[last slate]

no

[gusty falcon]

it's select all 2 element subsets from a pool of 11

[quiet grove]

i find it becomes easier with questions because u get the approach better... hard qs tho

[wooden badger]

But you said uve been taught combinations

[wooden badger]

Ill take the compliment

[golden ore]

Seriously, which grade?

[wooden badger]

11th bro tf

Do i look old

[golden ore]

I don't judge people on their faces, c'mon.

And I'm in 12th.

[wooden badger]

Ik

[quiet grove]

u studying for jee bro?

[golden ore]

yeah.

[hallow basin]

Hey Lnrd, consider rewatch some youtube videos about combination & permutations

[quiet grove]

same

[wooden badger]

How did u judge then. Uni TT

[hallow basin]

It would help

[last slate]

but thats not how I imagine combination

[quiet grove]

how do u imagine combination

[last slate]

I imagine it permutation(n,r)/r!

[quiet grove]

thats the formula

[golden ore]

I assumed you were an University student from overseas, not gonna lie.

[wooden badger]

(I can never be a teacher)

[quiet grove]

what does the formula mean to you

[last slate]

yeah but how does this apply to this problem

[hallow basin]

You have to know what's the meaning of combination and permutation

[wooden badger]

Why, Thank you but wot basiS

[last slate]

I know

[quiet grove]

you wanted the feel of permutations of letters with repeated letters right

[hallow basin]

Lmao

[last slate]

yes

[quiet grove]

for that you need to know what the formulae mean

and what they do

to derive the expressoin

[hallow basin]

Then you should have no problem doing this question

[golden ore]

I thoguht this was a uni level discussion channel.

[last slate]

but i do

[golden ore]

So I just "assumed".

[last slate]

because its a different scenario than what im used to

[wooden badger]

Cant tell no more if thts sarcasm

[last slate]

bruuuuuuuyh

[wooden badger]

Hey leon what kindof ques can you solve?

[quiet grove]

tell me what you're aware of

[quiet grove]

yeah

[wooden badger]

Dumbledore said calmly

[last slate]

thats how i imagine it

[quiet grove]

what is the q in this

[last slate]

there isnt i made it up

[quiet grove]

what did you make up

[last slate]

heres a question though

[quiet grove]

okay yeah how would you do the 1st q

[last slate]

Thats P(12, 4) / 4!

[wooden badger]

That's 12c4

[quiet grove]

okay so you know that that gives you the number of ways of choosing 4 from 12 right

[last slate]

and thats 4! To 1

[last slate]

it gives you the number of solutions but aren’t duplicated

[quiet grove]

is that what they taught you

[last slate]

I guess?

[quiet grove]

thats right

[last slate]

i mean i figured it out like that

[quiet grove]

cool

in my opinion its easier to first use combination and then permutation tho

[last slate]

can we look at both ways

[quiet grove]

yeah

both are right

but i need to know if you know what combinations mean

[quiet grove]

then u can understand this example

[last slate]

but theyre doing it in multiple parts and i dont get it

i dont get what theyre doing man

[quiet grove]

no worries gang i can try and explain

but before that you first need to know what combinations are

[quiet grove]

in the first question here

[last slate]

yes

i understand that

[quiet grove]

they're asking the number of ways of choosing 4 people out of 12 people

so that is just 12C4

[last slate]

yes

[quiet grove]

similarly, if i asked you the number of ways of choosing 2 from 11, what would your answer be

[last slate]

but the 2 here are specific

In the other problem the 4 people could be random

[quiet grove]

wym specific

this is also random

any 2 people from 11 people

[last slate]

Yes but it isnt saying that in the missippi question

It doesnt say choose 2 random letters

from the word

[quiet grove]

why are u jumping bro ill get to that

[last slate]

and see how many times you can rearrange them without duplication

[last slate]

alright

11 C 2 yes

[quiet grove]

yeah

now coming to the image u sent

we arent choosing 2 letters out of 11

imagine 11 empty spaces

[last slate]

ok

[quiet grove]

we're choosing 2 empty spaces out of the 11

to put the Ps in

did u get me?

[last slate]

sure?

i mean you’re not looking at the letters then

which I thought we needed to

[quiet grove]

wdym

we first chose the number of ways to place the 2 Ps

now we'll go to the other letters

[wooden badger]

There 11 seats not 11 ppl...

Acc ima shutup

[last slate]

but i thought we are talking about permutations of the letters

but without their duplicates

[quiet grove]

okay ill do a simpler one that maybe could explain why we're using combinations for permutations

you know the number of ways to arrange 5 letters with themselves is 5!?

[last slate]

yes

i know

[quiet grove]

we can show that from combinations

take the word plate

the n umber of ways to place P would be 5c1 because we're choosing 1 space from 5 empty spaces

[last slate]

i mean i would just say 5

bruh

[quiet grove]

yeah

5c1 is 5

now u already chose a place for P

so there are 4 places left

number of ways of choosing a place for L is 4c1

which is 4

same way 3 then 2 then 1

so you get 5.4.3.2.1

which is 5!

now applying the same concept for MISSISSIPPI

instead of choosing 1 letter at a time

you take the all the letters which are same, then choose places to place them in

[last slate]

so youre doing calculations on the slots not the letters?

[quiet grove]

yeah

[wise flicker]

we still on this?

[quiet grove]

yea

we're doing this

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

Channel closed

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[small warren]

Can I get help solving this question? I used Desmos to make sure that my answer is correct because I felt it was correct but it turned out to be wrong, Desmos shows that the graph has a minimum at -4 y, but my process led to me having 46 as the y for the minimum. I used the discriminant formula to make sure that the original equation has 2 real solutions and the solution of b^2-4ac was greater than 0.

[small warren]

This is my work

When i graphed the last equation it had no real solutions, so i was weirded out and graphed the original equation and it showed -5,-4 as the vertex.

what did i do wrong?

[fallow scarab]

when you add 25 on the inside, you have to subtract 75 on the outside because of the 3

[small warren]

Ah!!!!

Thank youuuu ❤️

[fallow scarab]

ab = a(b + c) - ac

[small warren]

You're righttt

Tysmmm ❤️

i'm so silly

.close

[midnight plank]

Channel closed

Closed by @small warren

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.
• Type the command .close to free the channel when you're done.
• 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.

[small warren]

Can I get help with this problem? the last option says when x > 4 that the exponential function has the highest value, but plugging in 4.01 shows that the linear equation is bigger, however for whole integers such as 5 the exponential function is bigger

[small warren]

Here's my work with the third option that i disproved

The 2nd option i feel is intuitively false

[shell wigeon]

,calc 20(4.01) + 4

[grand pond]

Result:

84.2

[shell wigeon]

,calc 3^(4.01) + 2

[grand pond]

Result:

83.894782046966

[small warren]

Yeah see?

But usually there's 3 correct answers that they want me to pick, i think

[shell wigeon]

"usually"?

[small warren]

,calc 3^(8)+2

[grand pond]

Result:

6563

[small warren]

Yeah in my school

[fallow scarab]

don't be a conspiracy theorist

[small warren]

my online exams

[small warren]

no i mean there is actually also 2 answers sometimes

[shell wigeon]

I mean you just said that last option is false, and you're correct

[small warren]

Ty!

Tysmmm ❤️

.close

[midnight plank]

Channel closed

Closed by @small warren

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.
• Type the command .close to free the channel when you're done.
• 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]

can someone help

[last slate]

what are we counting here exactly?

<@&286206848099549185>

[coarse wing]

hi

[livid merlin]

Number of donuts and their varieties.

[coarse wing]

counting possible list of totals for each flavor

[livid merlin]

Try permutations and combinations.

[coarse wing]

the donuts are "indistinguishable" it doesnt matter which specific chocolate donut you grab. All that matters is how many of each type end up in the box

[bold peak]

Okay let's think of it this way

You want three numbers that add up to 8

All of them non negative

But also the order of the numbers matters

So (0,0,8) is different from (8,0,0) cuz different types of donuts

You get me?

And for the second, you're setting, let's say the third number to always be at least 3

[coarse wing]

xavier you are dominating the help channel :<<

[bold peak]

Sorry

All yours 🌺

[coarse wing]

thanks

[last slate]

i can use as many helpers possible

[last slate]

but like

My mind was programmed to think of what we are counting

does order matter here

[last slate]

idk bruh is it like 3^8 / 3!

im so confused bruh this isnt even a joke anymore

I actually lowkey am gonna pay someone if they help me understand this stuff

like im ready to go that rout

[bold peak]

👀

[bold peak]

Nope

It's a partition problem

Also not taking payment cuz I'm not good enough at combi to warrant that lol

[last slate]

combi?

[coarse wing]

sorry i was eating

:pp

No, the order of the donuts does not matter

[bold peak]

Ah welcome back

[last slate]

it’s okay

[bold peak]

All yours again

[coarse wing]

okay let me handle here

[last slate]

ive been stuck on this all freakin day

[coarse wing]

okay

[coarse wing]

this one incorrect thi

because

3^8 counts situations where you pick one of three flavors for each of the 8 slots, and the order matters

​3! is for arranging 3 unique items. We only use it here to eliminate the order of the 3 variables, which isn't what we need to do

do you get it?

[last slate]

so what are we teying to do tho

[last slate]

no bro

so far i know combinations and permutations

[coarse wing]

hmm

[coarse wing]

what is the difference between a sequence and a final count?

focus on this question

[last slate]

a sequence is a set of ordered n tuples i guess

a final count is the amount of those ordered things?

[coarse wing]

yes!

now if your box contains (5 Chocolate, 3 Glazed) donuts, and you had a different order of picks, say (3 Glazed, 5 Chocolate) on a different day, does the final content of your box change?

@last slate

[last slate]

no

it does not

[coarse wing]

exactly

[coarse wing]

now did you understand this?

since the order of the picks doesn't change the final count of the indistinguishable items, we are counting combinations with repetition, not sequences

[last slate]

Can you tell me what cartesian product were doing here

so then i know what we need to count

are we doing permutations

like how are we arranging our items

[coarse wing]

we arent doing simple permutations for the same reason: we arent arranging 8 unique items where the order of selection matters

[last slate]

so whats the arrangement

[coarse wing]

were arranging 8 identical items (the donuts) into 3 distinct piles (the flavors)

@last slate

[last slate]

But I thought we are creating 8tuples

from the 3 flavors

no?

[coarse wing]

what are we doing here for the whole time Combinations with Repetition

like what you said

okay lets give another one

[last slate]

so we arent making 8tuples

[coarse wing]

yes

[coarse wing]

the arrangement we are counting is not the order of the 8 selections, but the order of the 8 donuts and the 2 dividers

[last slate]

they want a set of 8 donuts tho

[last slate]

sorry i don’t understand

[coarse wing]

youre 100% correct, the final answer is a set of 8 donuts

however

let me think a way for you to be more easier to understand

lets say the easiest way to count how many different sets are possible

is to pretend you are arranging the items

im gonna use the trick (arrangement of 10 symbols ( * and | ) for this

[last slate]

But i just wanna know what the arrangement is like

we have 3 varieties of donuts

we want the possible combinations of all donuts in some way

[coarse wing]

now select 3 donuts (k=3) from 3 varieties (n=3)

[last slate]

Chocolate vanilla glaze

i cant bruh

i dont think i ever got this confused before

This is actually crazy

[coarse wing]

Lets see what the arrangement is like for that simple case: 3 donuts (*) and 2 dividers (|). Total symbols: 5.

If we arrange the 5 symbols like this: * | * | *

we got 1 Chocolate, 1 Glaze, 1 Vanilla (as you guessed)

[last slate]

okay?

whats the point of this way

[coarse wing]

now if we arrange the 5 symbols like this: * * * | |

the first section has 3 donuts. The second section has 0 donuts. The third section has 0 donuts

we get 3 choco, 0 glaze and 0 vanilla

see how the arrangement of the * and | directly creates the final set of counts?

@last slate

as we can see every unique arrangement of the 5 symbols creates a unique final set, all we have to do is count the total arrangements of those 5 symbols which is $\binom{5}{2} = 10$ possibilities

[grand pond]

linh

[last slate]

so you’re rearranging the slots

But why is it 3 donuts

theyres asking for 8 I don’t get it

[coarse wing]

the 3 donut example just to show you how the arrangement works and how we get to the answer of 10 for that simple case (k=3)

now apply the same method i said to the 8 donuts

@last slate

[last slate]

okay so

sorry i don’t understand man

[coarse wing]

okay

[last slate]

Something is just not clicking

[coarse wing]

we need 8 donuts (*). We need 2 dividers (|) to split them into the 3 flavors.
So we are arranging $8 + 2 = \mathbf{10}$ symbols total

[grand pond]

linh

[last slate]

i dont get why the dividers are part of the rearrangement

like why not just rearrange the donuts 😭

like they want donuts so my brain tells me to do cartesian product like 3 ^8

Because thats from 3x3x3x3x3x3x3x3 = {(a,b,c,d,e,f,g,h) where each letter is some flavor }

[coarse wing]

you know this?

[last slate]

like i don’t understand why this is wrong

[coarse wing]

$\text{Ways} = \binom{k + n - 1}{n - 1}$

[grand pond]

linh

[last slate]

sure? But like i dont get its logic

discrete math sucks

[coarse wing]

The dividers are necessary because the 8 donuts are indistinguishable (* is the same as *)

If you only arrange the 8 donuts, you have 8! / 8! = 1 way. This doesnt help

the dividers (|) are what create the three distinct categories (the x1, x2, x3 variables). Without the two bars, you cant tell the counts apart

[last slate]

@coarse wing can u please tell me why this is wrong

[coarse wing]

the logic is that we convert the hard problem of counting sets with repetition into the easy problem of counting simple combinations by creating 10 slots and choosing which 2 get the distinguishing elements (the dividers)

[coarse wing]

okay si

so*

3^8 counts ordered sequences (tuples). However, the donut problem asks for unordered sets

[last slate]

is that where you do k to 1 mapping

[coarse wing]

Because we only care about the final count {x1, x2, x3}, we must use the method that eliminates order

[last slate]

because you want a set where order doesnt matter right

[coarse wing]

exactly

[last slate]

okay can we please continue from here

alright so you wanna eliminate the duplicates right

so you know each item can be arrnaged 8! Times

So P(3,8)/8!

?

[coarse wing]

the correct way to eliminate duplicates for k items is to divide by k!. However, this only works if all k items were distinguishable in the first place (like dividing a permutation by k!)

since we have repetition, dividing by 8! does not correctly eliminate the duplicates

but you are getting close to understand it

as the formula i said: $\text{Ways} = \binom{k + n - 1}{n - 1}$

[grand pond]

linh

[coarse wing]

we know that the question one is 8 donuts and 3 varieties => k = 8 and n = 3

[last slate]

yeah so what do we divide by

i know something is different because we have multi sets

but

i cant seem to figure it out

[coarse wing]

http://www.youtube.com/watch?v=40HxI6Uc00Q

do you know this?

[coarse wing]

You dont divide by anything.
​You directly calculate the number of arrangements of the dividers

[last slate]

no tbh it made things more confusing

[last slate]

you know what i give up

[coarse wing]

I think you should study the theory again carefully, because if I explain it and you cant remember it, it will be very difficult

[last slate]

but the book doesn’t explain properly

where do you recommend that i learn it properly

[coarse wing]

here the answer for question 1
$\binom{8 + 3 - 1}{3 - 1} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = \mathbf{45}$

[grand pond]

linh

[coarse wing]

if you understand this then apply the same logic to question 2

[last slate]

i dont unfortunately

[coarse wing]

http://www.youtube.com/watch?v=40HxI6Uc00Q

take a look at this

then comeback to me later

since the stars and bars method is often poorly explained in textbookz

[midnight plank]

@last slate Has your question been resolved?

[last slate]

Whats the intuition behind stars choose bars

[midnight plank]

@last slate Has your question been resolved?

[mortal falcon]

you have n indistinguishable objects that you want to put into k buckets. instead of dealing with the buckets directly, we can imagine separators between buckets. then, we can order our objects and separators in a line, so that each object is put into a bucket based on which separators it's between

[coarse wing]

the intuition is that once you decide where the dividers (bars) go, the problem is solved - the stars (donuts) must fill the remaining spots

sorry for late reply

[midnight plank]

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[last slate]

its okay

[last slate]

can i get help on this problem

[wise flicker]

Btw Which region you from?

[last slate]

im so frustrated

Im gonna break my screen

[wise flicker]

Lmao

Why did you do that?

[last slate]

because those four characters can be arranged 4! Ways

[wise flicker]

Yes, but that is internal arrangement with respect to themselves

[last slate]

this is what they’re saying

i dont freakin understand

what the hell is goin on rn

Why P(18,4)

youre putting what into 4 permutations

18 what

like bruh

[wraith dirge]

P(32,14) is the number of ways to arrange the non a729 characters, P(18,4) is the number of ways to arrange the 4 characters in the 18 character length

[last slate]

hoppoooow

i get P(32,14)

[wraith dirge]

you understand the P(32,14) right

[last slate]

not P(18,4)

[last slate]

yes

[wise flicker]

There are 18 slots, you arrange A729 in 4 of them

Other 14 are arranged

[hallow crane]

If you think about it then choose 14 characters out of 32 so 32C14 then arrange all the 18 in random order. So 18! 32C14 this would give same result upon simplification

[last slate]

thats P(4,4) bro

[hallow crane]

18 places 4 no.s

[wraith dirge]

no, because those 4 characters can go in any of the 18 places

[coarse wing]

what is happening here

[wraith dirge]

idk

#❓how-to-get-help

[coarse wing]

@last slate

did you watch the vid i sent

.

[last slate]

yeah but i didnt get it

[coarse wing]

oh no the permutations and combinations again..

[coarse wing]

I think you have that knowledge gap

[last slate]

please tell me how the cartesian product looks like here

just tell me

[last slate]

but this stuff is before