Question

Two finite sets have ‘m’ and ‘n’ elements. If the total number of subsets of the first set is 56 more than the total number of subsets of the second set, then find the values of ‘m’ and ‘n’.

Answer 1

Hint: Use the fact that the number of subsets of a set having ‘x’ elements is ${{2}^{x}}$. Write an equation based on the data given in the question. Simplify the equation to calculate the number of elements present in each set.

Complete step-by-step solution -
We know that two finite sets have ‘m’ and ‘n’ elements each and the number of subsets of the first set is 56 more than the number of subsets of the second set. We have to calculate the values of ‘m’ and ‘n’.
We know that the number of subsets of a set having ‘x’ elements is ${{2}^{x}}$.
Thus, the number of subsets of a set having ‘m’ elements is ${{2}^{m}}$ and the number of subsets of a set having ‘n’ elements is ${{2}^{n}}$.
We know that the number of subsets of a set with ‘m’ elements is 56 more than the number of subsets of a set with ‘n’ elements. Thus, we have ${{2}^{m}}-{{2}^{n}}=56$.
We will now simplify this equation. Taking out ${{2}^{n}}$ from both terms on the left-hand side, we ${{2}^{n}}\left( {{2}^{m-n}}-1 \right)=56$.
We observe that ${{2}^{n}}$ and ${{2}^{m-n}}$ is an even number. So, ${{2}^{m-n}}-1$ is an odd number.
Thus, the only possible way to write 56 as a product of an odd and even number is $56=8\times 7$.
Thus, we have ${{2}^{n}}=8$ and ${{2}^{m-n}}-1=7$.
Simplifying the above equations, we have ${{2}^{n}}={{2}^{3}}=8$ and ${{2}^{m-n}}=7+1=8={{2}^{3}}$.
Thus, we have $n=3$ and $m-n=3$.
So, we have $m-3=3\Rightarrow m=3+3=6$.
Hence, the values of ‘m’ and ‘n’ are $m=6,n=3$.

Note: We can’t solve this question without using the fact that the number of subsets of a set having ‘x’ elements is ${{2}^{x}}$. We need to use the trick to rearrange the terms to figure out the factors of 56 which will satisfy the given equation.