This is INSANE !!

$P_n = \binom{n-2}{3}$ and $Q_n = \tfrac13n\binom{n-4}{2}$ , so that $P_n - Q_n = n-4$ , and hence $X = 6$ .

If $f(x) = \sum_{r=0}^n\binom{n}{r}\frac{x^r}{r+2}$ , then $\tfrac{d}{dx}\big(x^2f(x)\big) =x(1+x)^n$ , from which we obtain $\sum_{r=0}^n \frac{\binom{n}{r}}{n^r(n+2)} \; = \; f(\tfrac{1}{n}) \; = \; \frac{(n+1)^n}{n^n(n+2)} + \frac{n^2}{(n+1)(n+2)}$ and hence $Y = 1 +2 + \tfrac{2}{2} = 4$ .

The function $\left\lfloor \tfrac{(x-2)^3}{m} \right\rfloor$ has fewer than $5$ discontinuities in $[4,6]$ for $m > \tfrac{64}{5}$ , $5$ discontinuities in $[4,6]$ for $\tfrac{64}{6} < m \le \tfrac{64}{5}$ , $6$ discontinuities in $[4,6]$ for $\tfrac{64}{7} < m \le \tfrac{64}{6}$ , and $7$ or more discontinuities in $[4,6]$ for $m < \tfrac{64}{7}$ . The only zero of $\sin(x-2)$ in $[4,6]$ is $x = 2 + \pi$ . Thus we deduce that $\left\lfloor \tfrac{(x-2)^3}{m}\right\rfloor\,\sin(x-2)$ has exactly $5$ discontinuities in $[4,6]$ when $m$ belongs to the set $\left\{ \tfrac13\pi^3\right\} \cup \left(\tfrac{64}{6},\tfrac{64}{5}\right]$ (when $m = \tfrac13\pi^3$ one of the 6 points of discontinuity of $\left\lfloor \tfrac{(x-2)^3}{m} \right\rfloor$ in $[4,6]$ occurs at $x = 2+\pi$ , and so is not a discontinuity of $\left\lfloor \tfrac{(x-2)^3}{m}\right\rfloor\,\sin(x-2)$ . Thus $A = \tfrac13\pi^3$ and $B = \tfrac{64}{5}$ , making the answer $\tfrac{5}{32} \times \tfrac{64}{5} + \left\lfloor \tfrac{100}{3}\pi^3\right\rfloor \; = \; 2 + 1033 = \boxed{1035}$