Combinatorial Proof of Sum of Binomial Coefficients $\sum_{m=k}^{n-k}\binom{m}k\binom{n-m}k=\binom{n+1}{2k+1}$

Question

I was curious if anyone might be able to give me a hint as to how one might show the following identity combinatorially:

$$\sum_{m=k}^{n-k}\binom{m}k\binom{n-m}k=\binom{n+1}{2k+1}$$

For the left hand side I reason that we are choosing, for example, one object out of $1$ type, then choose from $4$ other types for another object. Continuing on we take $1$ object from $2$ types and then select another object from the remaining $3$ types. The right hand side seems to be telling us something about picking an odd number of objects, which makes me think my initial attempt might have been off the mark and there is instead I should focus on parity. Perhaps I'm missing something simple?

I don't understand why this question has been marked as duplicate, since the index here is m... — Adren, Oct 03 '17 at 21:41
This seems to have been closed with a wrong target. This would be better. — Arnaud D., Dec 10 '19 at 15:17

score 0 · Accepted Answer · answered Apr 03 '13 at 02:19

0

HINT: Consider the task of choosing $2k+1$ numbers from the set $\{0,1,\dots,n\}$ of $n+1$ integers. Let the numbers chosen be $a_0,\dots,a_{2k}$, listed in increasing order. Set $m=a_k$, so that $k$ of the chosen elements precede $m$ and $k$ follow $m$.

answered Apr 03 '13 at 02:19

Brian M. Scott

616,228

I definitely see where I went wrong now, thanks! – 114 Apr 03 '13 at 02:30
@Stopwatch: You’re welcome! – Brian M. Scott Apr 03 '13 at 02:31

Combinatorial Proof of Sum of Binomial Coefficients $\sum_{m=k}^{n-k}\binom{m}k\binom{n-m}k=\binom{n+1}{2k+1}$

1 Answers1