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NO! That is the number of PERMUTATIONS, not COMBINATIONS. In a combination, the order does not matter so that 1234 is the same as 1432 or 3412 etc.
Assuming the 4 numbers are different, the correct answer is 15 comprising
4 1-digit combinations,
6 2-digit combinations,
4 3-digit combinations and
1 4-digit combination.
Another way to look at it is that the first number can be in a combination or not. With each of these possibilities, the second can be in or out - giving 2*2 = 4 ways so far. With each of these there are two options for the third giving 2*2*2 = 8 combinations so far and then the last number makes it 2*2*2*2 = 16. But one of these combinations contains none of the numbers - each one is not in. Leaving that one out gives the answer 15.
In general, the number of combinations of any size, from n distinct objects is 2n and if you exclude the null combination, it is 2n - 1.
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