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
We know that if a set X contains k elements, then the number of subsets of X are 2k.
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.
$\therefore 2^{m}-2^{n}=112$
$\Rightarrow 2^{n}\left(2^{m-n}-1\right)=2 \times 2 \times 2 \times 2 \times 7$
$\Rightarrow 2^{n}\left(2^{m-n}-1\right)=2^{4}\left(2^{3}-1\right)$
$\Rightarrow n=4$ and $m-n=3$
$\therefore m-4=3$
$\Rightarrow m=7$
Thus, the values of m and n are 7 and 4, respectively.
Hence, the correct answer is option (b).