Solution: 2017 Winter Final - 14

Author: Michiel Smid

Question

Let $n \geq 3$ be an integer. Consider a uniformly random permutation $a_1a_2 \dots a_n$ of the set $\{1,2,\dots,n\}$. Define the events

Which of the following is true?

(a)

The events $A$ and $B$ are not independent.

(b)

The events $A$ and $B$ are independent.

(c)

None of the above.

Let S be all possible permutations: $ |S| = n! $
Let's determine A
$a_n = 1$ is fixed
The other n-1 numbers can be placed in any order: $ (n-1)! $
$ |A| = (n-1)! $
$ Pr(A) = \frac{(n-1)!}{n!} = \frac{(n-1)!}{n \cdot (n-1)!} = \frac{1}{n} $
Let's determine B
First, we pick 2 numbers to be $a_1$ and $a_2$: $ \binom{n}{2} $
The other n-2 numbers can be placed in any order: $ (n-2)! $
$ |B| = \binom{n}{2} \cdot (n-2)! $
$ |B| = \frac{n!}{2! (n-2)!} \cdot (n-2)! $
$ |B| = \frac{n!}{2!} $
$ Pr(B) = \frac{n!}{2! \cdot n!} = \frac{1}{2} $
Let's determine $ A \cap B $
$a_n$ is set to $n$: 1
We choose 2 numbers to be $a_1$ and $a_2$: $ \binom{n-1}{2} $
The other n-3 numbers can be placed in any order: $ (n-3)! $
$ |A \cap B| = \binom{n-1}{2} \cdot (n-3)! $
$ |A \cap B| = \frac{(n-1)!}{2! (n-3)!} \cdot (n-3)! $
$ |A \cap B| = \frac{(n-1)!}{2!} $
$ Pr(A \cap B) = \frac{(n-1)!}{2 \cdot n!} = \frac{1}{2n} $

Now, let’s check for independence

$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $

$ \frac{1}{2n} = \frac{1}{n} \cdot \frac{1}{2} $

$ \frac{1}{2n} = \frac{1}{2n} $

Since the equation is true, A and B are independent