Question

If the total number of $m$-element subsets of the set $A=\{a_1,a_2,...,a_n\}$ is $k$ times the number of $m$ element subsets containing $a_4$ then $n$ is

• A. $(m-1)k$
• B. $mk$
• C. $(m+1)k$
• D. $(m+2)k$

Answer 1

The correct option is B $mk$

From set of $n$ element selecting a subset of $m$ element
$={}^n{C}_m$

Now, $a_4$ is already selected.
$\therefore$ Total number of sets which contains $a_4$ is ${}^{n-1}{C}_{m-1}.$

Now, it is given that
${}^n{C}_m=k\cdot {}^{n-1}{C}_{m-1}\Rightarrow \frac{n!}{m!(n-m)!}=k\cdot \frac{(n-1)!}{(m-1)!(n-m)!}\Rightarrow n=mk$