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

Discussion:

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

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

If $a_1 = 3$ 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  $\binom {6+2}{2}$ = 28

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

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

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