Answer

Correct Answer : c ) 47

The question is from the PnC topic of Mathematics.

Here, we must select three or more different numbers from 1, 2, 3, 4, 5, 6, 7, and 8 such that 3 and 5 are always included and 7 and 8 are never included.

Here, if we select any number of numbers, 3 and 5 are always there, two numbers are fixed.

Now, we have to find the subsets of the set {1, 2, 3, 4, 5, 6, 7, 8} with 3 and 5 always present.

Now, we have to select from the remaining numbers - 1, 2, 4, 6, 7 and 8

We have to find the possible subsets of {1, 2, 4, 6, 7, 8}

For each number there are two possibility, either it can be in the subset or it cannot be in the subset.

So, total number of subset = 2⁶

This will include a null subset with no element, it will make selection of only two numbers, but we need three or more numbers.

Now, we are left with 2⁶ - 1 = 63

Now, in these 63 subsets, we also have those subsets where 7 and 8 both are there.

Let's fix 7 and 8 in the subset, now, we are left with {1, 2, 4, 6}

Out of these, we can have 2⁴ subsets = 16 subsets.

So, the required number is = 63 - 16 = 47.

Hence, (c) is the correct answer.