Let $\mathrm{X}={\left({ }^{10}{\mathrm{C}}_1\right)}^2+2{\left({ }^{10}{\mathrm{C}}_2\right)}^2+3{\left({ }^{10}{\mathrm{C}}_3\right)}^2+\dots +10{\left({ }^{10}{\mathrm{C}}_{10}\right)}^2$ where${ }^{10}{\mathrm{C}}_{\mathrm{r}},\mathrm{r}\in \{1,2,\dots ,10\}$ denote binomial coefficients. Then the value of $\frac{1}{1430}\mathrm{X}$ is ____________.

$\begin{array}{c}\mathrm{X}=\sum _{\mathrm{r}=1}^{10} \mathrm{r}\cdot {\left({ }^{10}{\mathrm{C}}_{\mathrm{r}}\right)}^2=\sum _{\mathrm{r}=1}^{10} \mathrm{r}{\cdot }^{10}{\mathrm{C}}_{\mathrm{r}}{\cdot }^{10}{\mathrm{C}}_{\mathrm{r}}\\ =10\cdot \sum _{\mathrm{r}=1}^{10}{ }^9{\mathrm{C}}_{\mathrm{r}-1}{\cdot }^{10}{\mathrm{C}}_{10-\mathrm{r}}=10.\mathrm{cofficient} \mathrm{of} {\mathrm{x}}^9 \mathrm{in} {(1+\mathrm{x})}^9{(1+\mathrm{x})}^{10}=10.\mathrm{cofficient} \mathrm{in}{(1+\mathrm{x})}^{19}=10{\cdot }^{19}{\mathrm{C}}_9\end{array}$

Now, $\frac{\mathrm{X}}{1430}=\frac{10{\cdot }^{19}{\mathrm{C}}_9}{1430}=\frac{{ }^{19}{\mathrm{C}}_9}{143}=\frac{{ }^{19}{\mathrm{C}}_9}{11\times 13}$

                   $=\frac{19\times 17\times 16}{8}=19\times 34=646$