Finding probabilities using permutations and combinations

Intro

Howdy!  I’m Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we’re going to learn how to find probabilities using permutations and combinations.  Here’s our problem statement: A corporation must appoint a president, chief executive officer, chief operating officer, and chief financial officer.  It must also appoint a planning committee with three different members.  There are 15 qualified candidates, and officers can also serve on the committee.  Complete Parts A through C below.

Part A

OK, Part A asks, “How many different ways can the officers be appointed?”  Well, to get this value, we need to calculate either a permutation or a combination.  And since the scientific calculator in my computer is not equipped to calculate permutations and combinations, I’ve gone out to a website called Calculators.net because it has this handy permutations and combinations calculator.  So we put in the total number for our set — we’re choosing among 15 qualified candidates, so I’ll put 15 in here.  And then the amount of each subset — that's how many people we have to select for.  So how many positions do we have here?  A president, CEO, COO, and CFO — so that's 1, 2, 3, 4 positions.  We’ve got four positions, so I calculate that.

And here are my values, but do I take the permutation, or do I take the combination?  Well, the question we need to ask ourselves is “Do we have any order that matters?  Is order important to what we’re trying to calculate?”  We’re calculating this for the officers that are going to be appointed.  The officers have a hierarchy, which is a type of order.  Therefore, order is important and so we’re going to select the permutation.  So there are, in this case, 32,760 different ways to appoint the officers.  Excellent!

Part B

Now, Part B asks, “How many different ways can the committee be appointed?”  The committee has 15 candidates that we’re choosing from, and we have three different members for the committee.  So we want three of 15 total.  And hear the order doesn't matter because were just choosing people for committee; once you’re on the committee, you're on the committee.  There's no indication that there's any order here with the committee.  So therefore that means we want to calculate a combination, which in this case you see here is 455.  Nice work!

Part C

And finally, Part C asks, “What is the probability of randomly selecting the committee members and getting the three youngest of the qualified candidates?”  Well, the probability is going to be the part over the whole, so we’ll put a fraction in here.  The part goes on top.  How many of the ways that you can pick qualified candidates represent the three youngest?  Well, there's only one way in which the three youngest members get on the committee.  How many ways are there total?  Well, for the committee we already calculated that there’s 455 different ways, so that’s going to be the whole.  I check my answer.  Fantastic!

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