Final answer:
The problem is related to permutations with replacement from a set S = {A, B, C, D}. All possible selections of two different members where order matters and repetition is allowed are listed, yielding a total of 16 different permutations.
Step-by-step explanation:
The question asks to list all the ways you can select two different members from the set S = {A, B, C, D}. Since order is important and repetition is allowed, we are essentially looking for permutations with replacement. This is a fundamental concept in probability and combinatorics.
To list all combinations, we pair each element in the set with every other element, including itself. The possible selections are:
- (A, A)
- (A, B)
- (A, C)
- (A, D)
- (B, A)
- (B, B)
- (B, C)
- (B, D)
- (C, A)
- (C, B)
- (C, C)
- (C, D)
- (D, A)
- (D, B)
- (D, C)
- (D, D)
There are a total of 4 × 4 = 16 possible selections, illustrating the permutations with replacement. This calculation follows the counting principle that establishes the total number of outcomes for two independent events is the product of the number of outcomes for each individual event, which in this case is the same set S.