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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 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.

For more problems: Pre-Regional Mathematics Olympiad Problems

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

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