Combinatorial Analysis

[Win_odd Dhamnekar]

Argue that there are exactly \(\displaystyle \binom{r}{k} \binom{n-1}{n-r+k} \) solutions of [math]x_1 + x_2 + ... + x_r = x_n[/math] for which exactly k of the \(\displaystyle x_i\) are equal to 0.

How to answer this question?

 

[Dr.Peterson]

Argue that there are exactly \(\displaystyle \binom{r}{k} \binom{n-1}{n-r+k} \) solutions of [math]x_1 + x_2 + ... + x_r = x_n[/math] for which exactly k of the \(\displaystyle x_i\) are equal to 0.

How to answer this question?

Please tell us what thoughts you have. You know better than to ask without showing any work at all.

Do you know how to answer it without the requirement of k zeros?

Have you thought about what the two factors of the claimed answer might mean?