**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

**Hence the answer is 120.**

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