In how many ways can we get a sum of at most 17 by throwing six distinct dice ? In how many ways can we get a sum greater than 17 ?

Explanation

Let `x_(1),x_(2),.., x_(6)` be the number that appears on the six dice.
According to the questions, `x_(1)+x_(2)+x_(3)+..+x_(6) le 17, " where " 1 le x_(i) le 6`
Now we can remove the inequality sign by introducing one dummy variable `x_(7)`, such that
`x_(1)+x_(2)+x_(3)+..+x_(6)+x_(7)=17, " where " x_(7) ge 0`
`therefore` No. of solutions of inequality (1) and equation (2) are same
`therefore` Required number of solutions,
= coefficient of `p^(17) " in " (p+p^(2)+..+p^(6))^(6)(1+p+p^(2)+..)`
=coefficient of `p^(17) " in " p^(6) (1+p+..+p^(5))^(6)(1+p+p^(2)+..)`
= coefficient of `p^(11) " in " (1+p+..+p^(5))^(6)(1+p+p^(2)+..)`
= coefficient of `p^(11) " in " ((1-p^(6))/(1-p))^(6)((1)/(1-p))`
= coefficient of `p^(11) " in " (1-6p^(6))(1-p)^(-7)`
`= ""^(17)C_(11)-6 ""^(11)C_(5)`.
Number of ways in which we can get a sum greater than 17
= total number of cases - number of cases in which the sum is at most 17
`=6^(6)-(""^(17)C_(11)-6^(11)C_(5))`