In how many ways can four persons, each throwing a dice once, make a sum of 13 ? 

 

Let $x_1,x_2,x_3$ be the numbers on the upper faces of the four dice. Then, the required number of ways is equal to the number of solutions of 

$x_1+x_2+x_3+x_4=13,\text{ where }1\le x_1,x_2,x_3,x_4\le 6$

The number of solutions of this equation is equal to the

Coefficient of $x^{13}\mathrm{in}{\left(x^1+x^2+\dots .+x^6\right)}^4$

= Coefficient of $x^9\mathrm{in}{\left(\frac{1-x^6}{1-x}\right)}^4$

= Coefficient of $x^9\text{ in }{\left(1-x^6\right)}^4(1-x{)}^{-4}$

Coefficient of $x^9\text{ in }\left({ }^4{C}_0{-}^4{C}_1{x}^6{+}^4{C}_2{x}^{12}+\dots \right)(1-x{)}^{-4}$

= Coefficient of $x^9{\text{ in }}^4{C}_0(1-x{)}^{-4}$

${=}^4{C}_0{\times }^{9+4-1}{C}_{4-1}{-}^4{C}_1{\times }^{3+4-1}{C}_{4-1}$

${=}^{12}{C}_3-4{\times }^6{C}_3=220-80=140$