$\sum _{k = 0}^ n \binom ka \binom {n-k}b = \binom {n+1}{a+b+1}$

Question

$\sum _{k = 0}^ n \binom ka \binom {n-k}b = \binom {n+1}{a+b+1} $

Can anyone please give me a hint to prove it using combinatorics ? I know that $n \geq a+b$ and $k$ will run from $a$ to $n-b$ . But I think it should be $\binom {n}{a+b} $.

Can anyone please help me?

My answer at the earlier question sketches a combinatorial proof. See also here. — Brian M. Scott, Oct 04 '20 at 08:22

cosmo5 · Answer 1 · 2020-10-04T08:25:23.790

1

Hint :

Consider choosing $a+b+1$ objects from a total of $n+1$ such that there are $a$ objects before and $b$ objects after the $(k+1)^{st}$ object.

edited Oct 04 '20 at 08:25

answered Oct 04 '20 at 08:17

cosmo5

10,629

Please explain more. I did not get you.@cosmo5 – anonymous Oct 04 '20 at 08:21
This is counting in two ways. LHS indicates picking two items at once. So we created a situation in which expression on LHS matches that on RHS. See other answers also, since this question is actually a duplicate. – cosmo5 Oct 04 '20 at 08:24

$\sum _{k = 0}^ n \binom ka \binom {n-k}b = \binom {n+1}{a+b+1}$

1 Answers1