Question

$\left(\genfrac{}{}{0ex}{}{30}{0}\right)\left(\genfrac{}{}{0ex}{}{30}{10}\right)-\left(\genfrac{}{}{0ex}{}{30}{1}\right)\left(\genfrac{}{}{0ex}{}{30}{11}\right)+\left(\genfrac{}{}{0ex}{}{30}{2}\right)\left(\genfrac{}{}{0ex}{}{30}{12}\right)+...+\left(\genfrac{}{}{0ex}{}{30}{20}\right)\left(\genfrac{}{}{0ex}{}{30}{30}\right)$ is equal to

• A. ${}^{30}{C}_{11}$
• B. ${}^{60}{C}_{10}$
• C. ${}^{30}{C}_{10}$
• D. ${}^{65}{C}_{55}$

Answer 1

Answer: (C)

Let $A=\left(\begin{array}{c}30\\ 0\end{array}\right)\left(\begin{array}{c}c30\\ 10\end{array}\right)-\left(\begin{array}{c}30\\ 1\end{array}\right)\left(\begin{array}{c}30\\ 11\end{array}\right)+\left(\begin{array}{c}30\\ 2\end{array}\right)\left(\begin{array}{c}30\\ 12\end{array}\right)-\dots +\left(\begin{array}{c}30\\ 20\end{array}\right)\left(\begin{array}{c}30\\ 30\end{array}\right)$
$\therefore A={}^{30}{C}_0\cdot {}^{30}{C}_{10}-{}^{30}{C}_1\cdot {}^{30}{C}_{11}+{}^{30}{C}_2\cdot {}^{30}{C}_{12}$
$-\dots +{}^{30}{C}_{30}{}^{30}{C}_{30}$
$=$ Coefficient of $x^{20}$ in $(1+x{)}^{30}(1-x{)}^{30}$
$=$ Coefficient of $x^{20}$ in ${\left(1-x^2\right)}^{30}$
$=$ Coefficient of $x^{20}$ in ${\sum }_{r=0}^{30}(-1{)}^{r^{30}}{C}_r{\left(x^2\right)}^r$
$=(-1{)}^{10}{}^{30}{C}_{10}$ [for coefficient of $x^{20}$, put $r=10$ ]
$={}^{30}{C}_{10}$