Question

Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second. The values of m and n are respectively

(a) 4, 7 (b) 7, 4 (c) 4, 4 (d) 7, 7

Answer 1

We know that if a set X contains k elements, then the number of subsets of X are 2^k.

It is given that the number of subsets of a set containing m elements is 112 more than the number of subsets of set containing n elements.

$$\begin{gathered} \therefore 2^m-2^n=112 \\ \Rightarrow 2^n\left(2^{m-n}-1\right)=2\times 2\times 2\times 2\times 7 \\ \Rightarrow 2^n\left(2^{m-n}-1\right)=2^4\left(2^3-1\right) \\ \Rightarrow n=4\mathrm{and}m-n=3 \\ \therefore m-4=3 \\ \Rightarrow m=7 \end{gathered}$$

Thus, the values of m and n are 7 and 4, respectively.

Hence, the correct answer is option (b).