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 of appears on the six dice.
The number of ways=number of solutions of inequation
`x_(1)+x_(2)+x_(3)+x_(4)+x_(5)+x_(6) gt 15`
here, `1 le_(i) le 6, i=1,2,3,4,5,6`
Total number of cases=`6^(6)=2^(6)xx3^(6)=64xx729=46656`
and number of ways to get the sum less than or equal to 15. which is 4501
Hence, the number of ways to get a sum greater than 15 is 46656-4501=42155