Question

If the number of $5$ elements subsets of the set $A\{a_1,a_2.....a_{20}\}$ of $20$ distinct elements is $k$ times the number of $5$ elements subsets containing $a_4$, then $k$ is

• A. $5$
• B. $\frac{20}{7}$
• C. $4$
• D. $\frac{10}{3}$

Answer 1

The correct option is C $4$

No of $5$ elements subset ${=}^{20}{C}_5=\frac{20!}{15!15!}=\frac{20\times 19\times 18\times 17\times 16\times 15!}{15!5\times 4\times 3\times 2\times 1}=15504$

No of $5$ elements subset containing $a_4$ ${=}^{19}{C}_4=\frac{19!}{4!15!}$

$\Rightarrow \frac{20!}{15!15!}=k\frac{19!}{4!15!}$

$\Rightarrow \frac{20\times 19!}{5\times 4!}=k\frac{19!}{4!}$

$\Rightarrow k=4$