# Permutation and Combinations to select a committee and election board

I am currently studying for the SOA exams. I have ran into a problem in Lecture Notes in Actuarial Mathematics - A Probability Course for the Actuaries - Marcel B. Finan that I thought was quite simple but my answer seems to be incorrect. The problem reads:

A newly formed hiking club has 25 members. Three members volunteered to serve on a 3-person executive committee that will consist of a president, vice president, and secretary. However, the members will be elected for the positions. An election committee of four will be created from the remaining 22 members. In how many ways can the club select its officers and election committee?

My immediate thought is simple combination mathematics: $$\text{# of ways to select officers and committee} = \binom{25}{3} \binom{22}{4} = 16824500$$ However, the answer key I was provided states that the correct answer is $$43890$$ and I am unsure how. After playing around with some numbers, I have determined that $$43890 = 6 \cdot \binom{22}{4} = 3!\cdot \binom{22}{4}$$ but I have no clue where this factor of $$3!$$ comes from rather than $$\binom{25}{3}$$. Am I missing something simple? Any direction or explanation is greatly appreciated.

I think you're misunderstanding what is intended by the problem statement. I blame this more on the problem statement more on you: their wording isn't very clear.

The problem statement actually means: 3 of the 25 people have volunteered for the president/VP/secretary jobs, but we haven't decided which of them will get which role yet. Then later we'll pick 4 people from the remaining 22 to form the election committee. That's why the answer is $$3!$$ (ways to decide who gets which role out of the president/VP/secretary) times $$\binom{22}{4}$$ (ways to pick 4 unordered names out of the remaining 22 candidates).

Extra note - even with your interpretation, I think you're still missing a factor of $$3!$$. The president/VP/secretary are individual named roles, so $$\binom{25}{3}$$ doesn't fully cover that choice: you still need to pick which person gets each individual role, which gives an extra $$3!$$.

Hey I know this is super late but in case anyone else runs into this question I have the answer.

Your assumption of the question is incorrect (don't worry it got me too) because you are not being asked to choose the original 3 members. They are all volunteers, so they are picking themselves. The only thing left in the air is the permutation of their office.

So 3P3 = 3! = 6

22C4 is correct at 7315.

And so (6)(7315) = 43890.

I hope this helps someone in the future.