Number of 8 digit numbers sum of whose digits is 4

Find the number of 8 digit numbers sum of whose digits is 4.

Discussion:

Suppose the number is $latex a_1 a_2 a_3 ... a_8 $.The possible values of $latex a_1 $ are 1, 2, 3, 4. We consider these four cases.

If $latex a_1 = 4 $ then all other digits are 0 (since sum of digits is 4). Hence there is only 1 such number.

If $latex a_1 = 3 $ then exactly one of the other 7 digits is 1. Hence there are 7 such numbers (depending on where the digit '1' is).

If $latex  a_1 = 2 $ then sum of the other seven digits is 2.

Hence we compute the number of non negative integer solutions of $latex  a_2 + ... + a_8 = 2 $ .

This equals  $latex \binom {6+2}{2} $ = 28

If $latex a_1 = 1 $ then sum of the other seven digits is 3.

Hence we compute the number of non negative integer solutions of  $latex a_2 + ... + a_8 $ = 3

This equals $latex \binom {6+3}{3} $ = 84

Hence the answer is 120.

For more problems: Pre-Regional Mathematics Olympiad Problems

Pre-Regional Mathematics Olympiad - 2012 - Problem 17 - Video