$ \sum_{x=1}^n {{n-1} \choose {x-1}} {N \choose x} = {{N+n-1}\choose n}$

Question

I am working on a random variable problem. To prove a PMF to be a valid one, I need to establish the following identity :

I tried proving it by induction and using factorials, but was unable to solve. Somebody please suggest how to proceed.

P.S.: According to my calculations for the random variable problem, this identity must hold. However I may be wrong and it may be the case that the identity is invalid. If so, please mention.

@Prometheus Thanks for pointing out. I've corrected the statement. — Megha, Nov 22 '21 at 08:17

score 4 · Accepted Answer · answered Nov 22 '21 at 08:16

4

Notice that $$\binom{n-1}{x-1}=\binom{n-1}{n-x}$$ by symmetry. Hence, you wish to solve $$\sum_{x=1}^n\binom{n-1}{n-x}\binom Nx=\binom{N+n-1}{n}.$$

This is known as Vandermonde's identity. You will find many solutions on this site.

answered Nov 22 '21 at 08:16

Zuy

4,656

Okay. I didn't know about Vandermonde's identity. Thank you. @Zuy – Megha Nov 22 '21 at 08:19

$ \sum_{x=1}^n {{n-1} \choose {x-1}} {N \choose x} = {{N+n-1}\choose n}$

1 Answers1