Question

If $\left({}^{21}{C}_r{+}^{21}{C}_{r-1}\right)$ (${}^{21}{C}_{21-r}{+}^{21}{C}_{22-r}$) $=(231{)}^2$ then value of 'r' can be

• A. $17$
• B. $18$
• C. $1‘9$
• D. $20$

Answer 1

Answer: (D)

$20$

$\left({}^{21}{C}_r\right)+\left({}^{21}{C}_{r-1}\right)\left({}^{21}{C}_{21-r}\right)({}^{21}{C}_{21-r}+{}^{21}{C}_{22-r})=(231{)}^2$

$\left({}^{22}{C}_r\right)\left({}^{22}{C}_{22-r}\right)={231}^2$

$\left({}^{22}{C}_r\right)\left({}^{22}{C}_{22-r}\right)={231}^2$

$\left({}^{22}{C}_r\right)\left({}^{22}{C}_r\right)={231}^2$

$\Rightarrow {}^{22}{C}_r=231$

$\frac{22!}{r!(22-r)!}=231$

$\Rightarrow \frac{\mathrm{22.21.20}!}{r!(22-r)!}=\frac{22.21}{2}$

$20!.2!=r!(22-r)!$

$\Rightarrow r=2$ or $r=20$

Hence $r=20$.