Story proof for an "Even-Odd" combinatorial identity

Question

I'm practicing the strategy of double counting a set ("story proof") to prove combinatorial identities. I came across this identity refered as "Even-Odd Identity" on Art of Problem Solving and had little progress. $$\sum_{k=0}^m (-1)^k \binom{n}{k} = (-1)^m \binom{n-1}{m}$$ (https://artofproblemsolving.com/wiki/index.php/Combinatorial_identity)

Here's my thinking so far:

The alternating signs hints at Inclusion-Exclusion principle. If we temporaily assume $m$ to be even then we have to find a smart choice of sets $A_n$ such that $\bigcup A_n$ gives all possible ways of choosing an $m$ people committee from $n-1$ people. But I didn't managed to find one which gives the binomial coefficients above.

Answer 1

Let me assume that $m$ is even, for the moment. We can then rewrite your identity without any minus signs like this: $$ \left[\binom n0+ \binom n2+\dots+\binom nm\right]=\left[\binom n1+\binom n3+\dots+\binom n{m-1}\right] + \binom{n-1}{m}. $$ This rewritten identity can be proven using a bijective proof. Here are the combinatorial interpretations for each side of the equation.

• LHS: Subsets of $\{1,\dots,n\}$ whose cardinality is at most $m$, and whose cardinality is even.

• RHS: Subsets of $\{1,\dots,n\}$ whose cardinality is at most $m$, and whose cardinality is odd, OR subsets of $\{1,\dots,n\}$ with cardinality $m$ which do not include $1$.

To prove the identity, we just need to give a bijection between these combinatorial sets. To do this, given sets $A$ and $B$, let $A \oplus B$ denote the symmetric difference of $A$ and $B$. This means $A\oplus B=(A\setminus B)\cup (B\setminus A)$.

Given a subset $S$ of $\{1,\dots,n\}$, with even cardinality at most $m$, there are two cases.

• If either $1\in S$ or $|S|<m$, then the corresponding set to $S$ is $S\oplus \{1\}$. Note that $S\oplus \{1\}$ will still have cardinality at most $m$, and the cardinality of $|S\oplus \{1\}|$ will be odd. Therefore, $S\oplus \{1\}$ is a set counted by $\left[\binom n1+\binom n3+\dots+\binom n{m-1}\right]$.

• If both $1\notin S$ and $|S|=m$, then the corresponding set to $S$ is simply $S$ itself. In this case, the image of $S$ is one of the sets counted by $\binom{n-1}m$.

To prove this is a bijection, it suffices to give the inverse mapping. Given $T$ with odd cardinality at most $m$, the corresponding set is $T\oplus \{1\}$. Given $T$ with cardinality $m$ such that $1\notin T$, the corresponding set is just $T$. It is routine to prove that these two mappings are inverse to each other.

When $m$ is odd, an analogous proof works.