A club has 14 members. In how many ways can a president, vp, and treasurer be chosen if two specified club members refuse to serve together. What I have so far. There are ${14 \choose 2}$ ways to choose a pair that will not serve together. If two members refuse to serve together, then there are 12 remaining club members to place into 3 distinguishable positions. This can be done in ${\frac {12!}{(12-3)!}}$ ways. Multiply ${14 \choose 2}$ . ${\frac {12!}{(12-3)!}}$ = 120,120. 
 A: Order three of them as president, VP, and treasurer ($_{14}P_3$) and subtract out the arrangements that have those two people (let's call them Statler and Waldorf).  There are $6$ possible ways that Statler and Waldorf could fill the offices, and $12$ ways apiece to fill the third office from the remaining pool.
So, $14 \cdot 13 \cdot 12 - 72 = 2112$.
A: 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.
