The total number of subsets of {1, 2, 6, 7} are?

16

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8

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64

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32

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Solution

The correct option is C $16$
We have to find the total number of subsets of ${1, 2, 6, 7}$ .
We know that, for a set containing $n$ elements, the total number of subsets is $2^{n}$ .
Consider ${1, 2, 6, 7}$ , wich has $4$ elements.
$∴$ here $n = 4$
Hence total number of subsets is $2^{4} = 16$ .
Thus the total number of subsets of ${1, 2, 6, 7}$ is $16$ .

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8, 7, 16, 5, 32, 3, 64, 1, 128, (?)

Q. The value of

\frac{18^{3} + 7^{3} + 3 \cdot 18 \cdot 7 \cdot 25}{3^{6} + 6 \cdot 243 \cdot 2 + 15 \cdot 181 \cdot 4 + 20 \cdot 27 \cdot 8 + 15 \cdot 9 \cdot 16 + 6 \cdot 3 \cdot 32 + 64}

Find the value of
$\frac{(18^{3} + 7^{3} + 3 \cdot 18 \cdot 7 \cdot 25)}{3^{6} + 6 \cdot 243 \cdot 2 + 15 \cdot 81 \cdot 4 + 20 \cdot 27 \cdot 8 + 15 \cdot 9 \cdot 16 + 6 \cdot 3 \cdot 32 + 64}$

The value of $\frac{(18^{3} + 7^{3} + 3.18.7.25)}{3^{6} + 6.243.2 + 15.81.4 + 20.27.8 + 15.9.16 + 6.3.32 + 64}$ is

\frac{8}{2^{2} + 1} + \frac{16}{2^{6} + 1} + \frac{32}{2^{14} + 1} + \frac{64}{2^{14} + 1} + \frac{128}{2^{18} + 1} =

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