In how many ways can we get a sum greater than 15 by throwing six distinct dice?

Explanation

Let `x_(1),x_(2),x_(3),x_(4),x_(5) and x_(6)` be the number that appears on the six dice.
The number of ways=Number of solutions of the inequation
`x_(1)+x_(2)+x_(3)+x_(4)+x_(5) le 15`
Introducing a dummy variable `x_(7)(x_(7)ge0)`, the inequation becomes an equation
`x_(1)+x_(2)+x_(3)+x_(4)+x_(5)+x_(6)+x_(7)=15`
Here, `1 le x_(i) le 6 ` for `i=1,2,3,4,5,6 and x_(7) ge0`
therefore, numer of solutions
=Coefficient of `x^(15)` in `(x+x^(2)+x^(3)+x^(4)+x^(5)+x^(6))^(6)xx(1+x+x^(2)+ . ..)` [neglecting higher powers]
`=.^(15)C_(9)-6xx.^(6)C_(3)=.^(15)C_(6)-6xx.^(9)C_(3)`
`=5005-504=4501`