Solution: 2018 Winter Final - 17

Author: Michiel Smid

Question

Let $n \geq 1$ be an integer. Consider a uniformly random permutation of the set $\{1,2,3,\dots,2n\}$. Define the event

A = "both the first element and the last element in the permutation are even integers".

What is $\Pr(A)$?

(a)

$\frac{n}{2(2n-1)}$

(b)

$\frac{2(2n-1)}{n-1}$

(c)

$\frac{n-1}{4n}$

(d)

$\frac{n-1}{2(2n-1)}$

Solution

Let S be the total number of ways to choose a permutation from a set of size $2n$: $2n!$

For the first element and the last element in the permutation are even integers, we have to choose 2 even numbers. There are two ways to rearrange the 2 even numbers: $\binom{n}{2} \cdot 2$

For the remaining positions in the permutation, we have $(2n-2)!$ ways to rearrange these numbers: $(2n-2)!$.

$Pr(A) = \frac{|A|}{|S|} = \frac{\binom{n}{2} \cdot 2 \cdot (2n-2)!}{2n!} = \frac{ \frac{n!}{2!(n-2)!} \cdot 2 \cdot (2n-2)!}{2n!} = \frac{n(n-1)}{2n(2n-1)} = \frac{n-1}{2(2n-1)}$