Q117 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
Two finite sets have '$m$' and '$n$' number of elements respectively. The total number of subsets of the first set is 112 more than the total number of subsets of the second set. Then the values of $\mathrm{m}$ and $\mathrm{n}$ are respectively.
- A7, 7
- B4, 4
- C4, 7
- Correct answerD7, 4
Solution
The number of subsets of a set with $m$ elements is $2^{m}$.
The number of subsets of a set with $n$ elements is $2^{n}$.
Given the condition $2^{m} - 2^{n} = 112$.
Factoring out $2^{n}$, we get $2^{n}(2^{m-n} - 1) = 112$.
Expressing $112$ as a product of a power of $2$ and an odd number: $112 = 16 \times 7 = 2^{4} \times 7$.
Comparing $2^{n}(2^{m-n} - 1) = 2^{4} \times 7$, we have $2^{n} = 2^{4}$, which implies $n = 4$.
Also, $2^{m-n} - 1 = 7$, which implies $2^{m-n} = 8 = 2^{3}$.
Thus, $m - n = 3$. Substituting $n = 4$, we get $m - 4 = 3$, so $m = 7$.
The values are $m = 7$ and $n = 4$.
Answer: 7, 4
The number of subsets of a set with $n$ elements is $2^{n}$.
Given the condition $2^{m} - 2^{n} = 112$.
Factoring out $2^{n}$, we get $2^{n}(2^{m-n} - 1) = 112$.
Expressing $112$ as a product of a power of $2$ and an odd number: $112 = 16 \times 7 = 2^{4} \times 7$.
Comparing $2^{n}(2^{m-n} - 1) = 2^{4} \times 7$, we have $2^{n} = 2^{4}$, which implies $n = 4$.
Also, $2^{m-n} - 1 = 7$, which implies $2^{m-n} = 8 = 2^{3}$.
Thus, $m - n = 3$. Substituting $n = 4$, we get $m - 4 = 3$, so $m = 7$.
The values are $m = 7$ and $n = 4$.
Answer: 7, 4
Hence, the correct answer is option: 'D'.