## ParthKohli one year ago Hey.

1. ParthKohli

Why 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$\prod_{i=1}^{n} \sum_{k=a_i}^{b_i}x^{k}$

2. imqwerty

number of solutions =$x_{i}$?

3. ParthKohli

Number of solutions = ordered pairs.

4. ParthKohli

@ganeshie8

5. ParthKohli

yeah

6. ParthKohli

ah, kinda makes sense to me now.

7. imqwerty

ahhgh that given expression is so bad..

8. ParthKohli

I 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 $$x^{20}$$

9. ParthKohli

$\left(x^{-3}\cdot \left(\frac{1 - x^{30}}{1 - x}\right)\right)^{3}$

10. ParthKohli

^ how do we calculate the coefficient of $$x^{20}$$ in that?

11. imqwerty

i didn't get that 2nd step? (x^-3 +x^-2..x^26)^3

12. ParthKohli

ah, 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 $$-3 \le x_i \le 26$$

13. ParthKohli

now refer to the above identity

14. imqwerty

:o ok now i get the ques :)

15. ParthKohli

but how do we calculate the coefficient? :(

16. ParthKohli

ah, since $$1 -x^{30}$$ cannot generate $$x^{20}$$ we can remove that and find the coefficient of $$x^{20}$$ in $$x^{-9} \cdot \dfrac{1}{(1-x)^3}$$

17. ParthKohli

I'm not sure about the above step... can you confirm it?

18. imqwerty

wait what was that equation where did it go? http://prntscr.com/8nkh3q

19. ParthKohli

maybe there are displaying problems... refresh?

20. ganeshie8

generating functions