Blackjack with Dice (Part II)

The number $N_n$ of ways in which five distinguishable dice can show a total of $n$ is the coefficient of $x^n$ in the polynomial $\begin{aligned} \big(x + x^2 + x^3 + x^4 + x^5 + x^6\big)^5 & = \; \left(\frac{x(1-x^6)}{1-x}\right)^5 \; = \; x^5(1 - x^6)^5(1-x)^{-5} \\ & = \; x^5(1 - 5x^6 + 10x^{12} - \cdots)(1-x)^{-5} \end{aligned}$ and hence, since $(1 - x)^{-5} \; = \; \sum_{n \ge 0} \binom{n+4}{4}x^n \hspace{2cm} |x| < 1$ we have $N_n \; =\; \left\{ \begin{array}{lll} \binom{n-1}{4} & \hspace{1cm} & 5 \le n \le 10 \\ \binom{n-1}{4} - 5\binom{n-7}{4} & & 11 \le n \le 16 \\ \binom{n-1}{4} - 5\binom{n-7}{4} + 10\binom{n-13}{4} & & 17 \le n \le 22 \end{array}\right.$ which is enough to calculate $\sum_{n=5}^{21}N_n \; = \; \sum_{n=0}^{16} \binom{n+4}{4} - 5\sum_{n=0}^{10}\binom{n+4}{4} + 10\sum_{n=0}^4\binom{n+4}{4} \; = \; 6594$ making the desired probability $\frac{6594}{6^5} \; = \; \frac{1099}{1296}$ giving the answer $1099 + 1296 = \boxed{2395}$ .