The number of ways to choose 3 elements from a set of 6 is given by the combination formula C(n, r) = n! / (r!(n-r)!). Here, C(6, 3) = 20.

The total number of subsets of D is 2^4 = 16. Proper subsets are all subsets except the set itself, so there are 16 - 1 = 15 proper subsets.

The total number of subsets is 2^4 = 16. Proper subsets exclude the set itself, so 16 - 1 = 15.

A subset must contain elements from the original set. {4} is not a subset of D.

The proper subsets of D = {1, 2} are {∅}, {1}, and {2}. So, there are 3 subsets, but excluding D itself gives us 2 proper subsets.

The power set of D = {1, 2} is {∅, {1}, {2}, {1, 2}} which contains 4 subsets.

A subset can only contain elements from the original set. {3} is not a subset of D.

A subset can only contain elements from the original set. {1} is not a subset of D.

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

The subsets containing 'a' are {a}, {a, b}, so there are 2 subsets.

E · F = 1*1 + 1*2 + 1*3 = 1 + 2 + 3 = 6.

E · F = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

E · F = 2*1 + (-1)*2 + 3*0 = 2 - 2 + 0 = 0.

E · F = 5*1 + 5*2 + 5*3 = 5 + 10 + 15 = 30.

E · F = x*2 + y*3 + z*4 = 10 gives the equation 2x + 3y + 4z = 10.

If 1 is included, we can choose from the remaining 4 elements (2, 3, 4, 5). The number of subsets is 2^4 = 16.

The total number of subsets is 2^4 = 16. Proper subsets exclude the set itself, so there are 16 - 1 = 15 proper subsets.

The power set of a set with n elements has 2^n elements. Here, n = 2, so 2^2 = 4.

The subsets containing 1 are {1}, {1, 2}, and the empty set is not included.

The total number of subsets is 2^4 = 16. Proper subsets exclude the set itself, so 16 - 1 = 15.

If 'a' is included, we can choose from the remaining elements {b, c}. The subsets containing 'a' are {a}, {a, b}, {a, c}, and {a, b, c}, totaling 4 subsets.

The union of two sets includes all unique elements from both sets. Here, the union is {a, b, c, d}.

The power set of a set with n elements has 2^n subsets. For E, n = 2, so the size of the power set is 2^2 = 4.

The prime numbers less than 10 are 2, 3, 5, and 7. Therefore, E = {2, 3, 5, 7}.

The prime numbers less than 10 are {2, 3, 5, 7}. The power set has 2^4 = 16 elements.

If 1 is included, we can choose from the remaining elements {2, 3, 4}, which has 2^3 = 8 subsets.

The total number of subsets of a set with n elements is 2^n. For F, n = 4, so there are 2^4 = 16 subsets. Proper subsets are total subsets minus the set itself, so 16 - 1 = 15.

The power set of a set with n elements has 2^n elements. For F, n = 3, so the power set has 2^3 = 8 elements.

The subsets containing exactly 2 elements are {a, b}, {a, c}, and {b, c}. So, there are 3 such subsets.

The even numbers between 1 and 10 are 2, 4, 6, and 8. Therefore, F = {2, 4, 6, 8}.