A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 15 members?

Question

 Permutations and Combinations

Answer 1

Start with the perspective of the role (President, VP, S-T):  How may possible Presidents are there?   15

Then, of the 14 people remaining (since the President can not hold two positions), how many people could be VP for each selection of President?   14

    That makes 15*14 so far.

Now, there are 13 people remaining who could be S-T for each pair of P-and-VP that we have selected.  So,

    15*14*13

 

Conveniently, in formula form, P(n,r) is n!/(n-r)!.

   P(15,3)  is  15!/(15-3)! is  15*14*13 when you cancel the 12! in the denominator.

   P(15,3) = 2730

Answer 2

Since each role has a different title, the order does matter (ex. - if Jim is P and Jill is VP, then that is not the same thing as Jill as P and Jim as VP). So we would use a permutation.

 

15P3

= 15!/(15-3)!

= 15!/12!

=15*14*13

=2730

 

Let's say the titles were all the same (rather than president and vice-president they just picked three cabinet members of equal rank instead. So if Jim & Jill were both elected to the cabinet, then it doesn't matter if they were first or second elected since they're getting the same role), then the order would no longer matter and you would use a combination.

 

15C3

= 15!/(3!*(15-3)!)

=15! / (3! * 12!)

= 15*14*13 / 3!

=455