Question

If ${}^{21}{C}_r{=}^{21}{C}_{r+1}$, then ${}^r{C}_3=$

Answer 1

We have,

${}^{21}{C}_r{=}^{21}{C}_{r+1}$

On solving the given relation, we get

$\frac{21!}{r!(21-r)!}=\frac{21!}{(r+1)r!(21-r-1)!}$

$\frac{1}{(21-r)(21-r-1)!}=\frac{1}{(r+1)(21-r-1)!}$

$\Rightarrow r+1=21-r$

$\therefore r=11$

Now,

${}^{11}{c}_3=\frac{11!}{3!\left(8\right)}$

$=\frac{11\times 10\times 9}{6}=165$

Hence, this is the answer.