MathDB
Counting number of ways to choose same number of lights

Source: Spain Mathematical Olympiad 2021 P5

May 10, 2021
combinatoricsVandermonde identityCombinatorial IdentitySpain

Problem Statement

We have 2n2n lights in two rows, numbered from 11 to nn in each row. Some (or none) of the lights are on and the others are off, we call that a "state". Two states are distinct if there is a light which is on in one of them and off in the other. We say that a state is good if there is the same number of lights turned on in the first row and in the second row.
Prove that the total number of good states divided by the total number of states is:
357(2n1)2nn! \frac{3 \cdot 5 \cdot 7 \cdots (2n-1)}{2^n n!}