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ParthKohli
 one year ago
Hey.
ParthKohli
 one year ago
Hey.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Why is the number of solutions to the following equation:\[x_1 + x_2 + \cdots + x_k=n\]given \(a_i <x_i < b_i \) equal to the coefficient of \(x^{n}\) in the expansion of\[\prod_{i=1}^{n} \sum_{k=a_i}^{b_i}x^{k}\]

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1number of solutions =\[x_{i} \]?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Number of solutions = ordered pairs.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0ah, kinda makes sense to me now.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1ahhgh that given expression is so bad..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I understood why. Now as an example if we want to calculate solutions of\[x_1 + x_2 +x_3=20\]subject to \(x_i \ge 3\) then\[(x^{3} + x^{2} + \cdots + x^{26})^3\]coefficient of \(x^{20}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(x^{3}\cdot \left(\frac{1  x^{30}}{1  x}\right)\right)^{3}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0^ how do we calculate the coefficient of \(x^{20}\) in that?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1i didn't get that 2nd step? (x^3 +x^2..x^26)^3

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0ah, the lower restriction is given to us: \(x_1, x_2, x_3 \ge 3\) since their sum is \(20\), the max. value any of them can take is \(26\) (if and only if the other two are 3) thus \(3 \le x_i \le 26\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0now refer to the above identity

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1:o ok now i get the ques :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0but how do we calculate the coefficient? :(

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0ah, since \(1 x^{30}\) cannot generate \(x^{20}\) we can remove that and find the coefficient of \(x^{20}\) in \(x^{9} \cdot \dfrac{1}{(1x)^3}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure about the above step... can you confirm it?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1wait what was that equation where did it go? http://prntscr.com/8nkh3q

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0maybe there are displaying problems... refresh?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0generating functions
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