Three dice are thrown. The probability of getting a sum which is a perfect square is :

Solution :
$$n\left( S \right) = 6 \times 6 \times 6.$$    Clearly, the sum varies from $$3$$ to $$18,$$  and among these $$4,\,9,\,16$$   are perfect squares.
The number of ways to get the sum $$4$$
$$=$$ the number of integral solutions of $${x_1} + {x_2} + {x_3} = 4$$    where $$1 \leqslant {x_1} \leqslant 6,\,1 \leqslant {x_2} \leqslant 6,\,1 \leqslant {x_3} \leqslant 6$$
$$\eqalign{ & = {\text{coefficient of }}{x^4}{\text{ in }}{\left( {x + {x^2} + ..... + {x^6}} \right)^3} \cr & = {\text{coefficient of }}x{\text{ in }}{\left( {\frac{{1 - {x^6}}}{{1 - x}}} \right)^3} \cr & = {\text{coefficient of }}x{\text{ in }}{\left( {1 - {x^6}} \right)^3}.{\left( {1 - x} \right)^{ - 3}} = {}^3{C_1} \cr} $$
Similarly, the number of ways to get the sum $$9$$
$$\eqalign{ & = {\text{coefficient of }}{x^6}{\text{ in }}{\left( {1 - {x^6}} \right)^3}.{\left( {1 - x} \right)^{ - 3}} \cr & = - 3 \times 1 + {}^8{C_6} \cr & = 28 - 3 \cr & = 25 \cr} $$
The number of ways to get the sum $$16$$
$$\eqalign{ & = {\text{coefficient of }}{x^{13}}{\text{ in }}{\left( {1 - {x^6}} \right)^3}.{\left( {1 - x} \right)^{ - 3}} \cr & = {\text{coefficient of }}{x^{13}}{\text{ in }}\left( {1 - 3{x^6} + 3{x^{12}} - {x^{18}}} \right).\left( {{}^2{C_0} + {}^3{C_1}x + {}^4{C_2}{x^2} + ....} \right) \cr & = {}^{15}{C_{13}} - 3 \times {}^9{C_7} + 3 \times {}^3{C_1} \cr & = 105 - 108 + 9 \cr & = 6 \cr & \therefore \,n\left( E \right) = 3 + 25 + 6 = 34 \cr & {\text{So, }}P\left( E \right) = \frac{{34}}{{6 \times 6 \times 6}} = \frac{{17}}{{108}}. \cr} $$