Binomial Theorem Ques 4
- $\binom{30}{0}\binom{30}{10}-\binom{30}{1}\binom{30}{11}+\binom{30}{2}\binom{30}{12}+\ldots+\binom{30}{20}\binom{30}{30}$ is equal to
$(2005,1 M)$
(a) ${ }^{30} C_{11}$
(b) ${ }^{60} C_{10}$
(c) ${ }^{30} C_{10}$
(d) ${ }^{65} C_{55}$
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Answer:
Correct Answer: 4.(c)
Solution: (c) Let $A=\binom{30}{0}\binom{30}{10}-\binom{30}{1}\binom{30}{11}+\binom{30}{2}\binom{30}{12}-\ldots+\binom{30}{20}\binom{30}{30}$
$ \therefore \quad A={ }^{30} C_0 \cdot{ }^{30} C_{10}-{ }^{30} C_1{ }^{30} C_{11}+{ }^{30} C_2{ }^{30} C_{12} $ $-\ldots+{ }^{30} C_{20}{ }^{30} C_{30} $
$=\text { Coefficient of } x^{20} \text { in }(1+x)^{30}(1-x)^{30} $
$=\text { Coefficient of } x^{20} \text { in }\left(1-x^2\right)^{30} $
$=\text { Coefficient of } x^{20} \text { in } \sum_{r=0}^{30}(-1)^{r^{30}} C_r\left(x^2\right)^r $
$\left.=(-1)^{10}{ }^{30} C_{10} \quad \text { [for coefficient of } x^{20}, \text { put } r=10\right] $
$={ }^{30} C_{10}$
BETA
