Proof by finding a bijection

Question

Explain why
$\sum_{i=0}^n {(-1)^i}{n \choose i} = 0$

I'm pretty sure know how to prove this using the Binomial Theorem:

$\sum_{i=0}^n {(-1)^i}{n \choose i}$

$ =(-1)^0{n \choose 0}+(-1)^1{n \choose 1}+(-1)^2{n \choose 2}+(-1)^3{n \choose 3}+ ... + (-1)^{n-1}{n \choose n-1}+(-1)^n{n \choose n}$

$={n \choose 0}-{n \choose 1}+{n \choose 2}-{n \choose 3}+...+(-1)^{n-1}{n \choose n-1}+(-1)^n{n \choose n} $

${n \choose 0}={n \choose n}, {n \choose 1}={n \choose n-1}$, etc etc.

All of these are cancelled out and you're left with 0.

But how would you do this proof through finding a bijection? Is it something like:

Let M be the set of all even numbers. Let N be the set of all odd numbers.

For every $m \in M$ there exist an $n \in N $ such that $M \rightarrow N$

Or is it related to how ${n \choose 0}={n \choose n}, {n \choose 1}={n \choose n-1}$?

I'm totally new to this type of proof so I'm completely at a loss, any help would be appreciated.

Answer 1

HINT: The positive terms in the sum are those for which $i$ is even, and the negative terms are those for which $i$ is odd. Thus, the sum of the positive terms is the number of subsets of $[n]=\{1,\ldots,n\}$ of even cardinality, and the absolute value of the sum of the negative terms is the number of subsets of $[n]$ of odd cardinality. Thus, you could prove it by finding a bijection between the subsets of $[n]$ of even cardinality and those of odd cardinality.

Note that this is possible only when $n>0$; for $n=0$ the theorem is false, since $\binom00=1$, not $0$.