I think is easier to explain it with a specific problem (but would be nice a generic answer):
We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?
Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$
To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?
I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 \choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N \choose k} = \frac{N!}{k!(N-k)!}$.