Find ${}^{n}C_{21}$ , If ${}^{n}C_{10} = {}^{n}C_{11}$

$1$

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$0$

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$11$

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$10$

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Solution

The correct option is A
$1$

Given:
${}^{n}C_{10} = {}^{n}C_{11}$
$\Rightarrow$ $\frac{n!}{10! (n - 10)!} = \frac{n!}{11! (n - 11)!}$
$\Rightarrow$ $11! (n - 11)! = 10! (n - 10)!$
$\Rightarrow$ $11 \times 10! (n - 11)! = 10! (n - 10) (n - 11)!$
$\Rightarrow$ $11 = n - 10$
$\Rightarrow$ $n = 11 + 10$
$\Rightarrow$ $n = 21$
$∴ {}^{n}C_{21} = {}^{21}C_{21}$
$= 1$
Hence, Option ‘A’ is Correct.

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