First slide

Introduction to sets

Question

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

Easy

A

$2^{n - 1}$
B

$2^{n}$
C

$2^{n + 1}$
D

$2^{2 n}$

Solution

Let the original set contains $(2 n + 1)$ elements, then subsets of this set containing more than n elements, i.e., subsets containing $(n + 1)$ elements, $(n + 2)$ elements,
$. . . . (2 n + 1)$ elements.
$∴$ Required number of subsets
$= {}^{2 n + 1}C_{n + 1} + {}^{2 n + 1}C_{n + 2} + . . . + {}^{2 n + 1}C_{2 n} + {}^{2 n + 1}C_{2 n + 1}$
$= {}^{2 n + 1}C_{n} + {}^{2 n + 1}C_{n - 1} + . . . + {}^{2 n + 1}C_{1} + {}^{2 n + 1}C_{0}$
$= {}^{2 n + 1}C_{0} + {}^{2 n + 1}C_{1} + {}^{2 n + 1}C_{2} + . . . + {}^{2 n + 1}C_{n - 1} + {}^{2 n + 1}C_{n}$
$= \frac{1}{2} [{(1 + 1)}^{2 n + 1}] = \frac{1}{2} [2^{2 n + 1}] = 2^{2 n} .$

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