4 bet selections with all possible doubles

Question

I have selected 4 football matches to bet on this evening and noticed I have the option to automatically place a bet on every possible double going. Betfair states that with 4 selections, you can place 6 doubles in total.

I think I'm along the right lines by saying that $\frac{4!}{(4-2)!}$ would give all the permutations possible, but as this gives a result of 12, I'm clearly missing something. I'm pretty sure it's something to do with the order of the combination of bets being unimportant, but am unsure of how to translate that.

Any ideas?

Answer 1

The expression $4!/(4-2)!$ is correct only if you want to calculate all possible permutations where the order matters (i.e., AB is different from BA). Generalizing, the number of ways to get an ordered subset of k elements from a set of n elements can be calculated as $n!/(n-k)!$, whereas in the case of unordered subset the formula is $n!/[k!(n-k)!]$.

Answer 2

So after a bit of trial and error, and some Googling, turns out the expression I was looking for was:

4!/(2! X (4-2)!) = 6