Three dice are thrown together. The probability that the sum of the numbers appearing on them is 9, is

$\frac{7}{54}$

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

Total number of ways of getting 9 as the sum

= Coefficient of  $x^9$ in $(x^1 + x^2+x^3+x^4+x^5+x^6)^3$

= Coefficient of $x^6$ in $(1+x + x^2+x^3+x^4+x^5)^3$

= Coefficient of  $x^6$ in $\left(\frac{1-x^6}{1-x}\right)^3$

= Coefficient of  $x^6$ in $(1-x)^{-3}= {^{6+3-1}C}_{3-1}= {^8C}_2= 28 $

∴ Favourable number of elementary events = 28.

So, required probability $=\frac{28}{216}=\frac{7}{54}$