I assume you want to know a permutation about how many ways they can be in the final set-up.
Order matters, which is why this is a permutation.
Formula: P(n, r) = [tex] \frac{n!}{(n-r)!} [/tex]
where the permutation is [tex] n^{P} k[/tex], with P for permutation.
Plug in:
4!/(4-2)!
4!/2!
(1 x 2 x 3 x 4)/1 x 2
(3x4)/1
= 12
12 ways.
