In how many ways can we get a sum of atmost $15$ by throwing six distinct dice ? 

Let $x_1 , x_2 , x_3 , x_4 , x_5$ and x6 be the number that appears on the six dice. 

The number of ways $=$ Number of solutions of the inequation

       $x_1+x_2+x_3+x_4+x_5+x_6\le 15$

Introducing a dummy variable $x_7(x_7 \ge 0),$the inequation becomes an equation

       $x_1+x_2+x_3+x_4+x_5+x_6+x_7=15$

Here, $1\le x_i\le 6$ for $i=1,2,3,4,5,6$ and $x_7\ge 0.$

Therefore, number of solutions

    $=$Coefficient of $x^{15}$ in ${\left(x+x^2+x^3+x^4+x^5+x^6\right)}^{6}x\left(1+x+x^2+\dots \right)$

    $=$Coefficient of $x^9$ in ${\left(1-x^6\right)}^6(1-x{)}^{-7}$

   $=$Coefficient of $x^9$ in $\left(1-6x^6\right)\left(1{+}^7{C}_{1}x{+}^8{C}_2{x}^2+\dots \right)$ [neglecting higher powers] 

  $\begin{array}{l}{=}^{15}{C}_9-6{\times }^9{C}_3{=}^{15}{C}_6-6{\times }^9{C}_3\\ =5005-504=4501\end{array}$