SAT Math Explained: Permutations
This week, we’re going to discuss a math topic that, at first, sounds scary and intimidating: permutations. We’ll soon see that this scary monster is really a cuddly, although not-yet-house-broken, Chihuahua that loves to have its belly rubbed.
Permutation asks about the number of possible arrangements, and the order of the arrangement matters. For easier permutation problems, we could use our calculator. For example, consider this permutation problem:
If a clothing store has 3 models and 5 sets of outfits and each model can wear 1 outfit, how many different arrangements are possible?
For that problem, we have 5 options and 3 spaces available. To solve the problem, go to the MATH button and, under Probability, find the “nPr” button. Then, type 5 nPr 3, where 5 is the number of options and 3 is the number of spaces available. We let the calculator do the heavy lifting and find our answer: 60 permutations. Piece of cake.
But the SAT writers don’t want to give you too many pieces of cake. Sometimes they will throw you a pork chop and have you chew on it for a while. It’s a little more work, but definitely worth it! Consider this problem:
Agatha, Bertie, Camilla, Deloris, and Eutyche are sitting on a park bench. If Camilla can’t sit on either end of the park bench, how many different arrangements are possible for the five girls?
This problem is more difficult. We have the added twist of Camilla, who can’t sit on either end. I’d like to know why she’s making things so difficult for us, but she hasn’t been answering my texts. How do we account for that change in the problem? Let’s try to break it down into simpler parts.
It will help to visualize the seats and the possible girls sitting in each seat, so we’ll draw spaces and fill those with numbers.
_____ _____ _____ _____ _____
We essentially have two permutations in one. First, we need to figure out the possibilities for the outside seats. Second, we need to figure out the possibilities for the 3 seats remaining inside. Let’s look first at the outside seats.
For the first seat, there are 4 possibilities for girls (each of them except Camilla).
For the last seat, there can only be 3 possibilities, since Camilla can’t sit there, and one of the girls has already chosen her seat.
Now, we can look at the 3 seats inside. There are 3 possible girls to sit in the first inside seat, since Camilla can sit here and 2 girls are already seated.
__3__ _____ _____
2 of the 5 girls can sit in the middle seat, since 3 are already sitting in the previous seats.
__3__ __2__ _____
And the final girl has nowhere else to sit but the last inner seat. She should have made up her mind more quickly!
__3__ __2__ __1__
Now that we have these arrangements, we can multiply them all together to find the total possibilities for this unique situation.
__4__ X __3__ X __2__ X __1__ X __3__ = 72
If Camilla doesn’t want to sit on the ends, the five girls have 72 possible arrangements. Because we broke the complex problem down into 2 easier permutations, we were able to arrive at the answer relatively simply. For the future, remember to break the complex problem down into simpler pieces. We looked at the outsides first, then the insides, then were able to solve the problem pretty easily. If you see this problem again trying to rain on your parade, you should laugh at its pathetic attempts to intimidate you and tear it up!