sum of the digits equals 5

Nice question.
We need to find integers between 0 to 9999, in which the sum of digits adds up to 5.
(1) One digit is 5 and all other are 0: 0005, 0050, 0500, 5000 or we can say that no. of ways we can arrange the digits = 4!/3! = 4 ways
(2)Three 1's and one 2: 1112, this can be done in 4!/3! = 4 ways
(3) One 4 and one 1: 4100, this can be done in 4!/2! = 12 ways
(4) One 3 and one 2: 3200, this can be done in 4!/2! = 12 ways
(5) One 3 and two 1's: 3110, this can be done in 4!/2! = 12 ways
(6) Two 2's and One 1: 2210, this can be done in 4!/2! = 12 ways

Therefore, required number of positive integers = (12 * 4) + (4 * 2) = 48 + 8 = 56

The correct answer is C.