Question

Let we are given set as $S=\left\{ 2,4,6,8,\ldots \ldots 20 \right\}$. What is the maximum number of subsets S have?
\[\begin{align}
  & A.10 \\
 & B.20 \\
 & C.512 \\
 & D.1024 \\
\end{align}\]

Answer 1

Hint: In this question, we are given a set S and we need to find its maximum number of subsets. For this, we will first calculate the number of elements in the set using arithmetic progression. Then we will use the formula given as the number of subsets of a set with n elements is equal to ${{2}^{n}}$. Formula for ${{n}^{th}}$ term in an AP is given by ${{a}_{n}}=a+\left( n-1 \right)d$ where ${{a}_{n}}$ is ${{n}^{th}}$ term, a is the first term and d is the common difference.

Complete step-by-step solution
Here we are given the set as $S=\left\{ 2,4,6,8,\ldots \ldots 20 \right\}$.
As we can see, it is in the form of series 2, 4, 6, 8 . . . . . . . . . . . . 20. Since the difference between the second term and the first term is similar to the difference between the third term and the second term. So we can say that it is an arithmetic progression with common difference “d” as 2.
Now, the first term of the AP denoted by 'a' is 2.
Here we need to find n if ${{a}_{n}}$ term is 20 (last term).
As we know, the formula of AP is given by ${{a}_{n}}=a+\left( n-1 \right)d$.
So putting in the values we get:
\[\begin{align}
  & 20=2+\left( n-1 \right)2 \\
 & \Rightarrow 20=2+2n-2 \\
 & \Rightarrow 20=2n \\
 & \Rightarrow n=10 \\
\end{align}\]
Hence 20 is the ${{10}^{th}}$ term.
So we can say that set S has 10 terms.
Now, we need to find the maximum number of subsets of the given set S having 10 elements. So we will find proper as well as improper subsets.
The number of subsets of a set having n elements is equal to ${{2}^{n}}$.
So the number of subsets of the given set having 10 elements is equal to ${{2}^{10}}$.
${{2}^{10}}$ can be written as $2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=1024$.
Hence 1024 is the required answer.
Hence option D is the correct answer.

Note: Students should note the improper subsets of a set A are two subsets which are $\varnothing $ and A itself. Proper subsets are the subsets except for these two subsets. Here we were required to find the maximum number of subsets so we had found proper as well as improper subsets. While calculating the number of elements, do not get confused between ${{a}_{n}}$ and n in the formula of AP. ${{a}_{n}}$ represents ${{n}^{th}}$ term of an AP.