In how many ways could a club of 13 members choose a president, vice president and treasurer

As I mentioned in the comments, the particular pair of people who will not serve together is known, so we do not need to choose a pair that will not serve together.

John has provided you with an Inclusion-Exclusion argument. Here is another method.

We consider cases.

Case 1: Neither of the two people who will not serve together is selected for an office.

That leaves us with twelve people from which to select the officers. There are twelve choices for president, eleven choices for vice-president, and ten choices for treasurer for a total of $12 \cdot 11 \cdot 10 = 1320$ ways to fill the office without using either member of the pair who will not serve together.

Case 2: Exactly one of the pair who will not serve together is selected to hold an office.

There are two ways to select exactly one member of the pair who will not serve together to hold office and three ways to select the office that person holds. That leaves us with twelve people who can fill the remaining two offices. They can be selected in $12 \cdot 11$ ways. Hence, there are $2 \cdot 3 \cdot 12 \cdot 11 = 792$ ways to select the officers if exactly one member of the pair who will not serve together holds an office.

Total: Since the two cases are mutually disjoint, the number of ways the offices can be filled if a particular pair of people will not serve together is $1320 + 792 = 2112$, as John found.

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In how many different ways can a president, vice-president, secretary and treasurer be selected from a club with 15 members?

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13 options for the president position; 12 options for the vice-president position, and 11 options for the secretary-treasurer position. In all, it gives 13*12*11 = 1716 possible ways to fill the offices. ANSWER Solved. ------------------- This problem is about PERMUTATIONS. On Permutations,  see introductory lessons

    - Introduction to Permutations

    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".