The sum $\sum _{r=1}^{10}\left(r^2+1\right)\times r!=m\times n!then$

$$\begin{gathered} \sum _{r=1}^{10}\left(r^2+1\right)\times r! \\ \begin{array}{l}=\sum _{\mathrm{r}=1}^{10} \left[\mathrm{r}\right(\mathrm{r}+1)-(\mathrm{r}-1\left)\right]\mathrm{r}!\\ =\sum _{\mathrm{r}=1}^{10} \left[\mathrm{r}\right(\mathrm{r}+1)!-(\mathrm{r}-1)\mathrm{r}!]=(1\times 2!-0\times 1!)+(2\times 3!-1\times 2!)\\ +\dots \dots \dots \dots \dots +(10\times 11!-9\times 10!)\\ =10\times 11!\end{array} \end{gathered}$$

$\Rightarrow m=10,n=11$