Two finite sets have m and n elements.


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 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).

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