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Find the number of integers between 1 and 1000 having the sum of thedigits 18. |
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Answer» Correct Answer - `.^(33)C_(18)-6xx.^(23)C_(8)` 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)` |
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