In how many ways te sum of upper faces of four distinct dices can be six.

Explanation

Here, the number of required will be equal to the number of solutions of `x_(1)+x_(2)+x_(3)+x_(4)=5` i.e., `1 le x_(i) le 6` for i=1,2,3,4.
Since, upper limit is 6, which is greater than required sum, so upper limit taken as infinite. So , number of sols is equal to coefficient of `alpha^(5)` in the expansion of
`(1+alpha+alpha^(2)+ . .+oo)^(4)`
=Coefficient of `alpha^(5)` in the exapansion of `(1-alpha)^(-4)`
=coefficient of `alpha^(5)` in the expansion of
`(1+.^(4)C_(1)alpha+.^(5)C_(2)alpha^(2)+ . .)`
`=.^(8)C_(5)=.^(8)C_(3)=(8*7*6)/(1*2*3)=56`