Let X be a set containing n elements. If two subsets A and B of X are picked at random, the probability that A and B have the same number of elements is
A.\[\dfrac{{{}^{2n}{C_n}}}{{{2^n}}}\]
B.\[\dfrac{1}{{{}^{2n}{C_n}}}\]
C.\[\dfrac{{1 \cdot 3 \cdot 5 \cdot \cdot \cdot \left( {2n - 1} \right)}}{{{2^n} \cdot n!}}\]
D.\[\dfrac{{{3^n}}}{{{4^n}}}\]

Hint: In this question, we need to determine the probability that A and B have the same number of elements such that A and B are the subsets of X which are picked at random out of n elements. For this, we will use the concept of total number of elements for a particular subset for n elements.

Complete step-by-step answer:
Given X is a set containing n elements, let the elements of the set X be
\[X = \left\{ {{a_1},{a_2},{a_3},.......{a_n}} \right\}\]
Now we know that the number of subset of a set containing n elements is \[{2^n}\], so the number of subset of the set X will be \[ = {2^n}\]
Now since the two subsets A and B of X are picked at random, so the total number of ways of choosing A and B will be \[ = {2^n} \times {2^n} = {2^{2n}}\]
Now it is said that subset A and B have the same number of elements, which means if the subset A has 2 numbers of elements then set B should also have 2 numbers of elements to it, so the required number of ways will be
\[{}^n{C_0}{}^n{C_0} + {}^n{C_1}{}^n{C_1} + {}^n{C_2}{}^n{C_2} + ........ + {}^n{C_n}{}^n{C_n} = {}^{2n}{C_n}\]
We know probability is the ratio of the required number of ways by the total number of ways of an event, hence we can write
\[P = \dfrac{{{}^{2n}{C_n}}}{{{2^{2n}}}}\]
Now by further solving this we can write
\[
\Rightarrow P = \dfrac{{\left( {\dfrac{{\left( {2n} \right)!}}{{\left( {2n - n} \right)!n!}}} \right)}}{{{2^{2n}}}} \\
   = \dfrac{{\left( {\dfrac{{\left( {2n} \right)!}}{{n!n!}}} \right)}}{{{2^{2n}}}} \\
   = \dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)......1}}{{n!n! \cdot {2^{2n}}}} \\
 \]
Now we take \[\left( {2n} \right)\]as the common term from the numerator, so we get
\[
\Rightarrow P = \dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)......1}}{{n!n! \cdot {2^{2n}}}} \\
   = \dfrac{{{2^n}\left\{ {1.3.5.7......\left( {2n - 1} \right).n!} \right\}}}{{n!n! \cdot {2^n} \cdot {2^n}}} \\
 \]
Hence by solving we get
\[P = \dfrac{{1.3.5.7......\left( {2n - 1} \right)}}{{n! \cdot {2^n}}}\]
Therefore, the probability that A and B have the same number of elements is\[ = \dfrac{{1.3.5.7......\left( {2n - 1} \right)}}{{n! \cdot {2^n}}}\]
So, the correct answer is “Option C”.

Note: The number of subsets of a set containing n elements is \[{2^n}\]. Subsets are the elements present in the set. It is essential that the subset should contain (include) all the elements only and only from the set from which it is defined.