Blackjack with Dice (Part I)

The number $N$ of different ways in which five distinct dice can sum to $21$ is equal to the coefficient of $x^{21}$ in the polynomial $\begin{aligned} \big(x + x^2 + x^3 + x^4 + x^5 + x^6\big)^5 & = \; \left(\frac{x(1 - x^6)}{1-x}\right)^5 \; = \; x^5(1 - x^6)^5(1 - x)^{-5} \\ & = \; x^5(1 - x)^{-5} - 5x^{11}(1 - x)^{-5} + 10x^{16}(1 - x)^{-5} - \cdots \end{aligned}$ Since $(1 - x)^{-5} \; = \; \sum_{n \ge 0}\binom{n+4}{4}x^n \hspace{2cm} |x| < 1$ we deduce that $N \; = \; \binom{20}{4} - 5\binom{14}{4} + 10\binom{8}{4} \; = \; 540$ which makes the desired probability $\frac{N}{6^5} \; = \; \frac{5}{72}$ so the answer is $5 + 72 = \boxed{77}$ .