Question

In how many ways can three persons, each throwing a single dice once, make a sum of $15$?

Answer 1

Numbers on the faces of the dice are $1,2,3,4,5,6$
$\therefore$ Required number of ways = coefficient of $x^{15}$ in $(x^1+x^2+x^3+x^4+x^5+x^6{)}^3$
$=\text{coefficient of}{x}^{15}in{x}^3(1+x+x^2+x^3+x^4+x^5{)}^3$
$=\text{coefficient of}{x}^{12}in(1+x+x^2+x^3+x^4+x^5{)}^3$
$=\text{coefficient of}{x}^{12}in(1-x^6{)}^3(1-x{)}^{-3}$
$=\text{coefficient of}{x}^{12}in(1-3x^6+3x^{12})(1+3x+6x^2+...+28x^6+...+91x^{12}+...)$
$=91-84+3=94-84$
$=10$