Find the number of integers between 1 and 1000 having the sum of the digits 18.

Answer

`.^(33)C_(18)-6xx.^(23)C_(8)`

Explanation

Any number between 1 and 100000 must be of less than six digits.
Therefore, it must be of the form `a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)`
where `a_(1),a_(2),a_(3),a_(4),a_(5),a_(6) in {0,1,2,..,9}`
Now given that `a_(1)+a_(2)+a_(3)+a_(4)+a_(5)+a_(6)=18`
where `0 le a_(i) le9, i=1,2,3,..,9`
`therefore` Required number of integers
=coefficient of `p^(18) " in" (1+p+p^(2)+..+ p^(9))^(6)`
= coefficient of `p^(18) " in" (1-p^(10))/(1-p))^(6)`
= coefficient of `p^(18) " in" [(1-p^(10))^(6)(1-p)^(-6)]`
=coefficient of `p^(18) " in" [(1- .^(6)C_(1)p^(10))(1-p)^(-16)]`
`= .^(33)C_(18)-6xx .^(23)C_(8)`