The number of subsets of a set with n elements is 2^n. Here, n = 5, so 2^5 = 32.

The number of subsets of a set with n elements is given by 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.

The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.

The number of subsets of a set with n elements is 2^n. Here, n=4, so the number of subsets is 2^4 = 16.

The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.

The only subset that does not contain y is {∅} and {x}. Total = 2.

The number of ways = 5C2 * 4C2 = 10 * 6 = 60.

The number of ways is C(6,2) * C(4,2) = 15 * 6 = 90.

The number of ways to choose 3 books from 7 is 7C3 = 7! / (3! * 4!) = 35.

The number of ways is C(8,3) = 56.

The number of ways to choose 3 from 5 is given by 5C3 = 10.

The number of ways is C(11, 3) = 165.

The number of ways to choose 3 letters from 11 distinct letters is 11C3 = 165.

Treat the 3 men as one unit. So, we have 3 units (MMM, W, W). Arrangements = 4! * 3! = 24 * 6 = 144.

The arrangements = 5! / (3! * 2!) = 10.

The arrangements = 6! / (3! * 2! * 1!) = 60.

The total arrangements = 6! / (3! * 2! * 1!) = 60.

The number of ways to choose 4 books from 10 is C(10, 4) = 210.

The number of arrangements is 4! = 24.

Each ball can go into any of the 3 boxes, so the total ways = 3^4 = 81.

The number of ways to choose 4 fruits from 10 is given by 10C4 = 210.

The number of ways to choose 4 letters from 11 distinct letters is 11C4 = 330.

The number of ways to choose 4 letters from 11 distinct letters is 11C4 = 330.

The number of ways to award 4 different prizes to 3 students is 3^4 = 81.

Each prize can go to any of the 3 students, so the total ways = 3^4 = 81.

The number of ways is C(10, 4) = 210.

The number of arrangements of 5 different books is 5! = 120.

The number of ways is C(10, 5) = 252.

Total arrangements = 5! = 120. Arrangements with the two letters together = 4! * 2! = 48. So, arrangements where they are not together = 120 - 48 = 72.

Treat the two specific letters as one unit. Then, we have 4 units to arrange: 4! = 24. The two letters can be arranged in 2! = 2 ways. Total = 24 * 2 = 48.