An identity involving the alternating sum of products of binomial coefficients

Question

I encountered this identity while working in a geometry problem, and I was wondering if there is a combinatorial proof of this too:

$$\sum_{k\ =\ 0}^{2j}\left(-1\right)^{k} \binom{n}{k}\binom{n}{2j - k} = \left(-1\right)^j \binom{n}{j} $$

Answer 1

Let $N$ be a set with size $n$. For each $k\in \{0,\dots,2j\}$, $\binom{n}{k}\binom n{2j-k}$ enumerates the set of ordered pairs $$ (A,B)\quad\text{such that}\quad A,B\subseteq N,|A|=k,|B|=2k-j $$ We are then summing over $k$, and adding in this count when $k$ is even, but subtracting when $k$ is odd. That is, the total universe, $U$, is the set of ordered pairs $(A,B)$ with $|A|+|B|=2j$, and we are counting some of these positively, and some of these negatively. That is, $$ \sum_{k=0}^{2j}\binom{n}{k}\binom n{2j-k}=\sum_{(A,B)\in U}(-1)^{|A|}\tag{$*$} $$

Let us define the following sign-reversing involution on this universe. I use $E\oplus F:= (E\setminus F)\cup (F\setminus E)$ to denote the symmetric difference of sets.

Given $(A,B)$, with $A,B\subseteq N,|A|+|B|=2j$, let $m$ be the smallest element of $A\oplus B$. (If $m$ does not exist, the involution is undefined). Then $f(A,B)$ is given by $(A\oplus \{m\},B\oplus \{m\})$.

Note that $f(f(A,B))=(A,B)$ for all $(A,B)$ in the domain of $f$, and $f$ has no fixed points. This means $f$ partitions its domain into pairs of the form $\{(A,B),f(A,B)\}$. Furthermore, the two ordered pairs within each of these pairs will have opposite parity for their first coordinate, so they cancel each other out in the summation in $(*)$. Since they do not affect the count, we can eliminate them, and deduce that $(*)$ is equal to the number of pairs $(A,B)$ which are not in the domain of $f$.

$f(A,B)$ is defined exactly when $A\oplus B\neq \varnothing$, or equivalently when $A\neq B$. This means that the only pairs being counted are those of the form $$ (A,A),\quad A\subseteq N,|A|=j, $$ and these are being counted with sign $(-1)^{|A|}=(-1)^j$. This is exactly described by $(-1)^j\binom nj$, completing the proof.

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These sign reversing involutions proofs are a common tool for giving combinatorial proofs of alternating sign summations. If you like, you can turn this into a direct bijective proof of the identity $$ \sum_{k\text{ even}}\binom{n}{k}\binom n{2j-k}=(-1)^j\binom nj +\sum_{k\text{ odd}}\binom{n}{k}\binom n{2j-k} $$