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

none of these

Total numbers of elementary events =$6^3$.

The sum of the numbers on three dice varies from 3 to 18 and among these 4, 9 and 16 are perfect squares.

∴ Favourable number of elementary events

= Coefficient of $x^4$ in $(x + x^2+...+x^6)^3$

+ Coefficient of $x^9$ in $(x+x^2+...+x^6)^3$

+ Coefficient of $x^{16}$ in $(x+x^2+...+x^6)^3$

$=3+25+6=34$

Hence, required probability $=\frac{34}{6^3}=\frac{17}{108}$