A club $\mathrm{N}$ with five members is shown below. $\mathrm{N}=\{$ Alfred, Blake, Cindy, Donald, Elaine $\}$ Tabbreviated as $N=\{A, B, C, D, E\}$ Assuming all members of the club are eligible, but that no one can hold more than one office, list and count the different ways the clyb could elect both a president and a treasurer. Choose the correct list of possible pairs of presidents and treasurers. A. $A B, A C, A D, A E, B A, B C, B D, B E, C A, C B, C D, C E, D A, D B, D C, D E, E A, E B, E C, E D$ B. $A A, B B, C C, D D, E E$ C. $A A, A B, A C, A D, A E, B A, B B, B C, B D, B E, C A, C B, C C, C D, C E, D A, D B, D C, D D, D E, E A, E B, E C, E D, E E$ D. $A B, A C, A D, A E, B C, B D, B E, C D, C E, D E$ How many ways can a president and treasurer be elected?

Step 1 ：The problem is asking for the number of ways to choose a president and a treasurer from a group of 5 people, where each person can only hold one position. This is a permutation problem, because the order in which the people are chosen matters (i.e., choosing person A as president and person B as treasurer is different from choosing person B as president and person A as treasurer).

Step 2 ：The formula for permutations is \(P(n, r) = \frac{n!}{(n-r)!}\), where n is the total number of items, r is the number of items to choose, and '!' denotes factorial. In this case, n = 5 (the total number of people) and r = 2 (the number of positions to fill).

Step 3 ：Substituting the values into the formula, we get \(P(5, 2) = \frac{5!}{(5-2)!} = 20\).

Step 4 ：So, there are 20 different ways to choose a president and a treasurer from a group of 5 people, where each person can only hold one position. This matches with option A in the question, which lists 20 different pairs of people.

Step 5 ：Final Answer: The correct list of possible pairs of presidents and treasurers is option A, and there are \(\boxed{20}\) ways to elect a president and a treasurer.