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$The value of \frac{\sum_{r = 2}^{10}^{r} C_{2} \cdot^{10} C_{r}}{10} is$

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Solution

$\begin{array}{l} \frac{1}{10} (\sum_{r = 2}^{10} \frac{r (r - 1)}{2} \cdot \frac{10!}{r! (10 - r)!}) \\ = \frac{1}{10} (\sum_{r = 2}^{10} \frac{1}{2} \cdot \frac{10!}{(r - 2)! (10 - r)!}) \\ = \frac{1}{10} (45 \sum_{r = 2}^{10} \frac{8!}{(r - 2)! (10 - r)!}) \\ = \frac{1}{2} (9 \sum_{r = 2}^{10}^{8} C_{r - 2}) \\ = \frac{1}{2} (9 \times (^{8} C_{0} +^{8} C_{1} + \dots . . +^{8} C_{8})) \\ = \frac{1}{2} \times 9 \times 2^{8} \\ = 1152 \end{array}$

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