Question

A set contains $(2n+1)$ elements. The number of subsets of this set containing more than $n$ elements is equal to

• A. $2^{n-1}$
• B. $2^n$
• C. $2^{n+1}$
• D. $2^{2n}$

Answer 1

Given a set contains $2n+1$ elements.
The number of subsets of the set containing more than $n$ elements is the sum of subsets of the sets containing the elements $(n+1)$, $(n+2)$....etc till $(2n+1)$.
The number of subsets containing the $n+1$ elements is ${}^{2n+1}{C}_{n+1}$.
Similar way, the number of subsets containing the elements $n+2$ is ${}^{2n+1}{C}_{n+2}$
continuing this process....
Finally, the number of subsets containing the elements $(2n+1)$ is ${}^{2n+1}{C}_{2n+1}$
Thus, the required number of subsets of the set containing more than $n$ elements is ${=}^{2n+1}{C}_{n+1}{+}^{2n+1}{C}_{n+2}..........{+}^{2n+1}{C}_{2n+1}$........(i)
To evaluate the above expression, use binomial expansion $(1+x{)}^{2n+1}{=}^{(2n+1)}{C}_0{+}^{(2n+1)}{C}_1(x\left){+}^{2n+1}{C}_2\right(x^2)......{.}^{(2n+1)}{C}_n(x^n).............{+}^{2n+1}{C}_{2n+1}(x^{2n+1})$
Applying the rule of ${}^n{C}_r{=}^n{C}_{n-r}$ and substituting $x=1$, we get
$(1+1{)}^{2n+1}{=}^{(2n+1)}{C}_{2n+1}{+}^{(2n+1)}{C}_{2n}{+}^{2n+1}{C}_{2n-1}......{.}^{(2n+1)}{C}_{n+1}..............{+}^{2n+1}{C}_{n+1}{}^{}{+}^{2n+1}{C}_{2n+1}$
$2^{2n+1}=2{[}^{2n+1}{C}_{n+1}{+}^{2n+1}{C}_{n+2}..........{+}^{2n+1}{C}_{2n+1}]$
$\frac{2^{2n+1}}{2}={[}^{2n+1}{C}_{n+1}{+}^{2n+1}{C}_{n+2}..........{+}^{2n+1}{C}_{2n+1}]$
Therefore, the required value is$=2^{2n}$

Hence, the correct option is B $\left(2^{2n}\right)$