I need combinatorial proof for the following expression

Question

$$\sum_{k=0}^{m}{{m \choose k}×{n \choose (p+k)}}$$

I tried so much but I couldn't find any good results

Answer 1

"I hope next time to show us you work"

$$\sum_{k=0}^{m-p} {n \choose k}{m \choose p-k} = {m+n-p \choose p}$$

To prove the expression $\sum_{k=0}^{m-p} {n \choose k}{m \choose p-k} = {m+n-p \choose p}$ combinatorially, we can consider a group of $m$ men and $n$ women, from which we want to select a team of $p$ men and any number of women up to $m-p$. We can count the number of ways to do this in two different ways:

1. First, we can choose the $p$ men from the $m$ available men in ${m \choose p}$ ways, and then choose any number of women up to $m-p$ from the $n$ available women in the sum of ${n \choose k}$ ways for $k$ ranging from 0 to $m-p$. This gives us the left-hand side of the given equation, which represents the total number of ways to select a team of $p$ men and any number of women up to $m-p$.

2. Alternatively, we can choose any number of women up to $m-p$ from the $n$ available women in the sum of ${n \choose k}$ ways for $k$ ranging from 0 to $m-p$, and then choose the remaining $p-k$ men from the remaining $m-(p-k)$ men in ${m-(p-k) \choose p-k}$ ways. This gives us the right-hand side of the given equation, which represents the total number of ways to select a team of $p$ men and any number of women up to $m-p$.

Since both expressions count the same thing (the number of ways to select a team of $p$ men and any number of women up to $m-p$), they must be equal. Therefore, we have proved the given expression combinatorially.

I hope I provided what is needed to you