MCQ
The value of $\left( \begin{array}{l}30\\0\end{array} \right)\,\left( \begin{array}{l}30\\10\end{array} \right) - \left( \begin{array}{l}30\\1\end{array} \right)\,\left( \begin{array}{l}30\\11\end{array} \right)$ + $\left( \begin{array}{l}30\\2\end{array} \right)\,\left( \begin{array}{l}30\\12\end{array} \right) + ....... + \left( \begin{array}{l}30\\20\end{array} \right)\,\left( \begin{array}{l}30\\30\end{array} \right)$
  • A
    $^{60}{C_{20}}$
  • $^{30}{C_{10}}$
  • C
    $^{60}{C_{30}}$
  • D
    $^{40}{C_{30}}$

Answer

Correct option: B.
$^{30}{C_{10}}$
b
(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 $(1 -x^2)^{30}={^{30}}{C^{10}}$.

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