The sum $\displaystyle\sum_{r =1}^{10} (r^2 + 1) \times (r!)$ is equal to :

Question

$\begin{array}{1 1}(1)\; (11)! \\(2)\; 10×(11!) \\ (3)\; 101×(10!) \\ (4)\; 11×(11!) \end{array}$