Let $X=\left({ }^{10} C_{1}\right)^{2}+2\left({ }^{10} C_{2}\right)^{2}+3\left({ }^{10} C_{3}\right)

Question

Let $X=\left({ }^{10} C_{1}\right)^{2}+2\left({ }^{10} C_{2}\right)^{2}+3\left({ }^{10} C_{3}\right)^{2}+\ldots+10\left({ }^{10} C_{10}\right)^{2}$, where ${ }^{10} C_{r}, r \in\{1,2, \ldots, 10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is

Professor PC Thomas Classes · Answer

(646) :\ $X=\sum_{r=1}^{10}\left({ }^{10} C_{r}ight)^{2}=\sum_{r=1}^{10} r \cdot{ }^{10} C_{r} \cdot{ }^{10} C_{r}$\ $=10 \cdot \sum_{r=1}^{10}{ }^{9} C_{r-1} \cdot{ }^{10} C_{10-r}=10 \cdot{ }^{19} C_{9}$\ $	herefore \quad \frac{X}{1430}=\frac{10 \cdot{ }^{19} C_{9}}{1430}=\frac{{ }^{19} C_{9}}{143}=646$\