> 
> Four players compete in a tournament and are ranked $1$ to $4$. They then compete in another tournament and are again ranked from $1$ to $4$. Suppose that their performances in the second tournament, so that the two sets of rankings are independent.
>  What is the probability that nobody receives the same ranking twice?
> 
> 
> 

One way to do it:

Let the notation $(2,3,1,4)$ represent the result that the player who finished 1st
in tournament 1 finished 2nd in tournament 2, the player who finished 2nd
in tournament 1 finished 3rd in tournament 2, the player who finished 3rd in
tournament 1 finished 1st in tournament 2, and the player who finished 4th in
tournament 1 finished 4th in tournament 2.

Total outcomes= $4! = 24$

The results where no player receives the same ranking in the two tournaments
are:
$$(2,1,4,3), (2,3,4,1), (2,4,1,3), (3,1,4,2), (3,4,1,2) (3,4,2,1), (4,1,2,3), (4,3,1,2),(4,3,2,1)$$
There are nine of these results and so the required probability is $9/24 = 3/8$

If anyone could give me another way of doing this question to give me an intuition into whats going on behind the numbers, it would help a lot.

Thank you!