A penny saved is a penny
Haskell, 37 34 bytes

s#l@(c:d)|s&gt;=c=(s-c)#l+s#d
s#_=0^s


Usage example: 26 # [1,5,10,25] -> 13.

Simple recursive approach: try both the next number in the list (as long as it is less or equal to the amount) and skip it. If subtracting the number leads to an amount of zero, take a 1 else (or if the list runs out of elements) take a 0. Sum those 1s and 0s.

Edit: @Damien: saved 3 bytes by pointing to a shorter base case for the recursion (which also can be found in @xnors answer).