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@ganeshie8 urgent...

\[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \]

That reduces to the coefficient of \(x^{20}\) in\[(x^{-3} + x^{-2} + \cdots + x^{20})^3\]

How do I calculate the above?

@dan815 use elementary techniques please

\[=\left( x^{-3}\cdot\dfrac{1-x^{30}}{1-x}\right)^3\]

\[= x^{-9}(1-x^{30})^3(1-x)^{-3 }\]

\[x^{-9} (1 - x^{90} - 3x^{30} + 3x^{60})(1-x)^{-3}\]

\[= (x^{-9} - x^{81}-3x^{21} + 3x^{51} )(1-x)^{-3}\]

now the coefficient of \(x^r\) in \((1 - x)^{-n}\) is\[\binom{n+r-1}{r-1}\]

you could also simply use stars and bars

hmm, yeah. is it \(y_1 + 3\) or \(y_1 - 3\)?

\(y_i=x_i+3\)
so yeah it should be \(y_i-3\)

ok, that's nice.\[\binom{n-1}{k-1}\]

watcha doing?

u want int solutions?

\[\binom{19}{2}\]

id just use stars and bars

but sure ill bite, how does it come out to be the coeff of x^k again

oh sorry, I missed all the threes.

\[y_1 + y_2 + y_3 = 29\]

should the number of nonnegative solutions be\[\binom{29-1}{3-1}\]

oh nvm i get whwy its the coeff of x^20 in that expansion kinda neat trick

yeah, it is.

the way to think about this problem is the same as
x1+x2+x3=k
x1,x2,x3 >= 0

do you know how to solve that problem

allowed -3 what u basicaaly did is just allow more 1s in each place

by introducing -3 you just shifted your number index

ik, ik. there are various tricks to find the coefficient.

ok so watcha looking for now

ya x1+x2+x3=29 is right

for x1,x2,x3 non negative solutions

ok, so nonnegative solutions|dw:1444026999220:dw|

hahahaha

eh, that thing was just meant to illustrate stars-and-bars for x1 + x2 + x3 = 4

ya i know

hey shut up

but yes parth

heres a very easy way to work it out

\[\binom{29+3-1}{3-1}\]

suppose
x1,x2,x3 = 20
but u allow no 0 x1,x2,x3>=1

what does this really mean?

that means we must start with 1 in x1 x2 x3

1 +k1+ 1 + k2 +1 +k3 = 20
so your new zero beginning equation is now
k1+k2+k3=17

now u see how we went from the 1 case to 0 case

same logic from 0 to -3 case

just number indices here they are just symbols

i know what's happening here jesus

dont sweat

swear

ohhh I got where I went wrong lol

the n?

Oh wait..

\[\binom{29+3-1}{3-1} \] this should be fixed no?

i need \((x^{-3} + x^{-2} + \cdots + x^{26})^3 \)

yay 465 is right baby

Yeah see it should be \[\binom{29+3-1}{29}\] ye?

well both of them are equal so it doesn't even matter

Ah I see, you're too good Mr. Kohli!

how many solutions to \(x+y+z+t=29\) when \(x\ge 1, y\ge 2, z\ge 3, t\ge0 \)

same thing

again we transform in the following manner\[x=a+1\]\[y=b+2\]\[z=c+3\]

yea

how many solutions to \(x_1 + x_2 + x_3 + 4x_4 = 20\)

coefficient of \(x^{20}\) in\[(1 + x + \cdots + x^{20})^3(1 + x^{4}+x^8+\cdots + x^{20}) \]

looks like you're back

so can you use that trick here? I haven't got the real feel for it.

\[\frac{\left(1 - x^{21}\right)^3 }{(1-x)^{3}} \cdot \frac{1- x^{20}}{1-x^{4} }\]

now can we remove \((1-x^{21})^3\) because it doesn't contribute to \(x^{20}\)?

\[(1-x)^{-3} \left(1 + x^4 +x^8 + x^{12}+x^{16}+x^{20}\right)\]

now the coefficient of \(x^r\) in \((1-x)^{-n}\) is\[\binom{n+r-1}{r-1}\]is that right?

or is it the opposite... coefficient of \(x^n\) in \((1-x)^{-r}\) is \(\binom{n+r-1}{r-1}\)

\[\frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k \]

\[ \frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{k }x^k \]

\[\dbinom{-n}{r} = (-1)^r\dbinom{n+r-1}{r-1}\]

that's nice

\[(1-x)^{-s}=\sum_{k=0}^{\infty} (-1)^k\binom{s+k-1}{s-1}x^k \]
right ?