# Combinatorial proof of binomial coefficient summation identity

I am trying to get a combinatorial proof for this summation identity, for $0\leq k\leq \frac{n}{2}$ $$\sum_{m=k}^{n-k}\binom{m}{k}\binom{n-m}{k}=\binom{n+1}{2k+1}$$

The way I interpreted is that, for the right hand side $\binom{n+1}{2k+1}$ is the number of ways to choose $2k+1$ from total of $n+1$, then for left hand side, condition on dividing $n+1$ into two groups of size $m+1$ and $(n+1)-(m+1)=n-m$, then if choose $k$ the group of $n-m$ then it gives $\binom{n-m}{k}$, so there are $2k+1-k=k+1$ to choose from the group of size $m+1$, this is $\binom{m+1}{k+1}$, therefore I got the left hand side as $\sum_{m=k}^{n-k}\binom{m+1}{k+1}\binom{n-m}{k}\neq\sum_{m=k}^{n-k}\binom{m}{k}\binom{n-m}{k}$. what did I do wrong here? please help.

Suppose that you pick a set $A$ of $2k+1$ members of the set $[n+1]=\{1,\dots,n+1\}$. Let $A=\{a_0,\dots,a_{2k}\}$, where $a_0<a_1<\ldots<a_{2k}$. Then $k$ members of $A$ are less than $a_k$, and $k$ are greater. Let $A_0=\{a_0,\dots,a_{k-1}\}$ and $A_1=\{a_{k+1},\dots,a_{2k}\}$. Clearly $A_0\subseteq[a_k-1]$, so there are $\binom{a_k-1}k$ possibilities for $A_0$. Similarly, $A_1\subseteq[n+1]\setminus[a_k]$, so there are $\binom{n+1-a_k}k$ possibilities for $A_1$. Thus, for a fixed value $\ell$ of $a_k$ there are

$$\binom{\ell-1}k\binom{n+1-\ell}k\tag{1}$$

possible $(2k+1)$-subsets $A$ of $[n+1]$. The minimum possible value of $\ell$ is $k+1$, since there must be at least $k$ smaller numbers in $[n+1]$, and the largest possible value is $n+1-k$. Let $m=\ell-1$; then $m$ ranges from $k$ through $n-k$, the expression $(1)$ becomes

$$\binom{m}k\binom{n-m}k\;,\tag{2}$$

and it’s now clear that summing $(2)$ from $m=k$ to $m=n-k$ is counting the $(2k+1)$-subsets of $[n+1]$ according to their middle elements.

In your approach you are in effect looking only at $A_0$ and $A_1$, but they don’t tell you where the cutoff between them is $-$ the $a_k$ in my construction.

• @user62453: You’re welcome! – Brian M. Scott Mar 19 '13 at 17:38
• You can avoid the switch from $\ell$ to $m=\ell+1$ by taking your initial $n+1$-set to be $\{0,1,\ldots,n\}$ rather than $\{1,2,\ldots,n+1\}$ (as I did in the answer I linked to). – Marc van Leeuwen Mar 22 '13 at 11:00

The way you are trying to set of things, there is no way to recover the value of $m$ from the subset chosen in the right hand side, so several choices of $m$ may lead to the same subset, and you're overcounting.

However if you take $m$ to be equal to the middle element of the selected subset taken in increasing order (that is the $k+1$-st smallest one), then you solve two difficulties: you can recover $m$ from the choice, and there remain onle $k$ elements to choose on either side of that value $m$. See this answer for handling a somewhat more general identity.