My head hurts

The total combinations of 4-digit no. to give 5 are:

0,0,0,5; 0,0,2,3; 1,1,1,2; 0,0,1,4; 0,1,1,3

Therefore total ways are : 4!/3! + 4!/2! + 4!/3! + 4!/2! + 4!/2!

= 4+12+4+12+12 = 56

1-99: 5, 14, 41, 23, 32, 50 100-199: the 100th digit is 1, it means the sum of the other two is 4, so these are the combinations of 04, 13, 22 = 2+2+1=5 200-299: the sum of the other two is 3, combinations of 03, 12 = 2+2= 4 300-399: the sum of the other two is 2, combinations of 02, 11 =2+1=3 400-499: the sum of the other two is 1 , combinations of 01 =2 500: 1 1000-1999: the sum of the three is 4, combinations of 004, 013, 022, 112= 3+6+3+3=15 2000-2999: the sum of the three is 3, combinations of 003, 012, 111 = 3+6+1=10 3000-3999: the sum of the three is 2, combinations of 002, 011= 3+3=6 4000-4999: the sum of the three is 1, combinations of 001 = 3 5000: 1

adding them all together= 6+5+4+3+2+1+15+10+6+3+1=56

Stars and bars approach: we have to separate 5 stars into 4 groups, the number of stars in each group representing the value of the corresponding digit in the result. This boils down to placing 3 bars between 5 stars in a row to separate them into 4 groups. So on 5 + 3 = 8 positions, we have to place 3 bars and 5 stars. The number of ways to do this is $\binom{8}{3} = 56$ .