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The value of $\left( \begin{array}{l}30\\0\end{array} \right)\,\left( \begin{array}{l}30\\10\end{array} \right) - \left( \begin{array}{l}30\\1\end{array} \right)\,\left( \begin{array}{l}30\\11\end{array} \right)$ + $\left( \begin{array}{l}30\\2\end{array} \right)\,\left( \begin{array}{l}30\\12\end{array} \right) + ....... + \left( \begin{array}{l}30\\20\end{array} \right)\,\left( \begin{array}{l}30\\30\end{array} \right)$
$^{60}{C_{20}}$
$^{30}{C_{10}}$
$^{60}{C_{30}}$
$^{40}{C_{30}}$
Solution
(b) ${(1 – x)^{30}} = {\,^{30}}{C_0}{x^0} – {\,^{30}}{C_1}{x^1} + {\,^{30}}{C_2}{x^2} + …… + {( – 1)^{30}}{\;^{30}}{C_{30}}{x^{30}}$….$(i)$
${(x + 1)^{30}} = {\,^{30}}{C_0}{x^{30}} + {\,^{30}}{C_1}{x^{29}} + {\,^{30}}{C_2}{x^{28}}+ …… + {\,^{30}}{C_{10}}{x^{20}} + …. + {\,^{30}}{C_{30}}{x^0}$….$(ii)$
Multiplying (i) and (ii) and equating the coefficient of $x^{20}$ on both sides, we get required sum
= coefficient of $x^{20}$ in $(1 -x^2)^{30}={^{30}}{C^{10}}$.
