Let $\displaystyle \mathrm{S}_{1}=\sum_{\mathrm{j}=1}^{10}j(j-1)^{10}\mathrm{C}_{\mathrm{j}}$ , $\displaystyle \mathrm{S}_{2}=\sum_{\mathrm{j}=1}^{10}j^{10}\mathrm{C}_{\mathrm{j}}$ and $\displaystyle \mathrm{S}_{3}=\sum_{\mathrm{j}=1}^{10}j^{2}10_{\mathrm{C}_{\mathrm{j}}}$.
Statement-1: $\mathrm{S}_{3}=55\times 2^{9}$
Statement-2: $\mathrm{S}_{1}=90\times 2^{8}$ and $\mathrm{S}_{2}=10\times 2^{8}$.

$<br />\mathrm{S}_{1}=\sum_{\mathrm{j}=1}^{10}|(j-1)\dfrac{10!}{(j-1)(j-2)!(10-j)!}=90\sum_{\mathrm{j}=2}^{10}\dfrac{8!}{(j-2)!(8-(j-2))!}=90\cdot 2^{8}<br />$

$<br />\mathrm{S}_{2}=\sum_{\mathrm{j}=1}^{10}|\dfrac{10!}{j(j-1)!(9-(j-1))!}=10\sum_{\mathrm{j}=1}^{10}\dfrac{9!}{(j-1)!(9-(j-1))!}=10\cdot 2^{9}<br />$

$\displaystyle \mathrm{S}_{3}=\sum_{\mathrm{j}=1}^{10}[j(j-1) +j]\displaystyle \dfrac{10!}{|!(10-j)!}=\sum_{\mathrm{j}=1}^{10}|(j-1)^{10}\mathrm{C}_{\mathrm{j}}=\sum_{\mathrm{j}=1}^{10}j^{10}\mathrm{C}_{\mathrm{j}}=90.2^{8}+10.2^{9}$

$=90.2^{8}+20.2^{8}=110.2^{8}=55.2^{9}$.