Question

Find ${}^n{C}_{21}$ if ${}^n{C}_{10}{=}^n{C}_{11}.$

• A. $1$
• B. $0$
• C. $11$
• D. $10$

Answer 1

Answer: (A)

Given, ${}^n{C}_{10}{=}^n{C}_{11}$
$\Rightarrow$ $\frac{n!}{10!(n-10)!}=\frac{n!}{11!(n-11)!}$
$\Rightarrow$ $11\times 10!(n-11)!=0!(n-10)(n-11)!$
$\Rightarrow$ $11=n-10$
$\Rightarrow$ $n=11+10=21$ Now, ${}^n{C}_{21}{=}^{21}{C}_{21}=1$