Question

The number of subsets of the set A = \left\\{ {{a_1},{a_2},...,{a_n}} \right\\} which contain even number of elements is
A) ${2^{n - 1}}$
B) ${2^n} - 1$
C) ${2^n} - 2$
D) ${2^n}$

Answer 1

Here, we will find the cardinality of the given set and then use that the total number of subsets of the given set is the cardinality of the power set of the same set. Then we have to divide the obtained cardinality of the power set by 2 to find the number of subsets, which contain even number of elements.

Complete step by step solution: We are given that the set is A = \left\\{ {{a_1},{a_2},...,{a_n}} \right\\}.

We know that the cardinality of a set is the number of elements in that set.

First, we will find the cardinality of the given set $A$.

$\left| A \right| = n$

Since we know that the total number of subsets of $A$ is the cardinality of the power set of $A$.

So, then we have $\left| {P\left( A \right)} \right| = {2^n}$.

Thus the total number of subsets of $A$ is ${2^n}$.

If the number of odd numbers is not zero, then half of the subsets of the odd numbers contain an even number in the given set.

Now we can find the number of subsets, which contain even number of elements by dividing the $\left| {P\left( A \right)} \right|$ by 2, we get

$$\Rightarrow \dfrac{{\left| {P\left( A \right)} \right|}}{2} \\\ \Rightarrow \dfrac{{{2^n}}}{2} \\\ \Rightarrow {2^{n - 1}} \\\$$

Therefore, the number of subsets of the given set, which contain even number of elements, is ${2^{n - 1}}$.

Hence, option A is correct.

Note: Since the number of subsets of even numbers, subsets of odd numbers, proper subsets and non-empty subsets are all different quantities and will have different answers for a given set. So we have to be really careful while finding the answer. Students should know the meaning of a set and its power set to understand this problem.