By story proof I mean proving a combinatorial identity by counting the number of elements of some carefully chosen set in two different ways to obtain the different expressions in the identity. The following is my favorite:
Prove (try the algebraic method!) $$\sum _ { k = 0 } ^ { n } {n \choose k} 2 ^ { k } { {n - k }\choose { \left[ \frac { n - k } { 2 } \right]} } = { {2 n + 1 } \choose { n }},$$
Proof: Assume there are $2n+1$ people, where one of them T is single (it's me), the rest of them are $n$ pairs of lovers $(a_n,b_n)$. Now we need to choose $n$ among the $2n+1$ people for a dance party. There are $2$ methods:
A. Choose $n$ people arbitrarily, which accounts for ${ {2 n + 1 } \choose { n }}$ combinations in total.
B. Fix $k$, where $0\leq k \leq n $. Choose $k$ pairs of lovers, and demand only one of them can go to the party, which accounts for ${n \choose k} 2 ^ { k }$ combinations. Choose ${ \left[ \frac { n - k } { 2 } \right] }$ from the remaining $n-k$ pairs, and the two of them don't need to be separated.
If $n-k$ is odd, $$ k + 2 \left[ \frac { n - k } { 2 } \right] = n - 1,$$ T can also join the party!
If $n-k$ is even, $$ k + 2 \left[ \frac { n - k } { 2 } \right] = n,$$ T cannot join the party :(
When $k$ is fixed, the total number of combinations of the $n$ people is ${ {n - k }\choose { \left[ \frac { n - k } { 2 } \right]} }$. Let $k$ vary from $0$ to $n$, then method B produces $\sum _ { k = 0 } ^ { n } {n \choose k} 2 ^ { k } { {n - k }\choose { \left[ \frac { n - k } { 2 } \right]} }$ combinations.
As the 2 methods should result in the same number of combinations, the identity is proved.
Source: My variant of Gu Jian's proof from Collection of CMO (Chinese Mathematics Olympiad) Problems (Sorry for can't provide a picture as it's written in Chinese).
I would like to collect everyone's favorite story proof. Apologize if this is a duplicate.
Blue's link: https://en.wikipedia.org/wiki/Double_counting_(proof_technique)
Any other less well-known ones?
