Positive integers less than 10000, so they must be four digit (or less than four digit) positive integers
Sum of digits equal to 5, those integers possibly have those sets of digits;
0,0,0,5
0,0,1,4
0,0,2,3
0,1,1,3
0,1,2,2
1,1,1,2
Then we calculate ways of Number can be formed from those sets using permutation. (three, two and one digit positive integers are also counted here, it's when one or more "zero" are put at the left positions - 0005 or 0023 for example)
0,0,0,5 = 4P4/3! (as "0" is used three times) = 4!/3! = 4
0,0,1,4 = 4P4/2! (as "0" is used twice) = 4!/2! = 12
0,0,2,3 = 4P4/2! (as "0" is used twice) = 12
0,1,1,3 = 4P4/2! (as "1" is used twice) = 12
0,1,2,2 = 4P4/2! (as "2" is used twice) = 12
1,1,1,2 = 4P4/3! (as "1" is used three times) = 4
Total ways = 12*4 + 4*2 = 56
Take C.

thanks buddy...nice explanation.