On the SAT, you’ll see a handful of permutation questions. Permutation questions deal with the rearranging of existing elements. Let’s look at an example problem. The problem goes something like this “A northeastern (museum/school/restaurant) has four (displays/desks/seats) all in a row. How many different combinations of six (paintings/students/diners) can be made?”
This kind of problem can seem difficult at the onset, but it is pretty straight forward when you get down to it. We start by thinking about each “space” as a number of possibilities.
If we have six distinct options, then any of the six could be in the first of four spaces. Thus, there are six possibilities for this first space. The second space is also six, right? Not so fast: one distinct option has already been used for the first space. Even though we don’t know which one is used, something is in that first space, so there are only five options left for the second space. This pattern continues for the next two spaces which contain four and three possibilities respectively. The final step is to simply multiply the numbers of possibilities together: 6 x 5 x 4 x 3 = 360, and voila we have our answer.
Be careful to check your work. In this question we are looking for 4 combinations with 6 possibilities, so we only multiply those 4 numbers (6 x 5 x 4 x 3).
This is a fairly straightforward question, but is very similar to many medium and even hard questions on the SAT. If we can think of each available “space” in terms of possibilities, we can attack even harder questions.
Let’s look at a more difficult example:
“Five people reflected by the letters A, B, C, D and E are arranged in a row. If either A or E has to be in the first or the last position, what is the probability that B will be in the second position?”
Quick review: Probability is the number of desired outcomes divided by the number of total outcomes, so those two numbers are all we need to find. So how many desired outcomes do we have? Well, if A is in the first position then B, C, or D could be in second, third or fourth and E must be last, and if E is in the first position then B, C, or D could be in second, third or fourth and A must be last, so let’s just write AB and EB to start and fill in the rest until we can’t any more. If we do this, our desired outcomes are ABCDE, ABDCE, EBCDA, and EBDCA. Four outcomes: easy enough.
So what are our total outcomes?
- Well how many possibilities do we have for our first position? Either A or D: Two options.
- How about our second? B, C, or D: Three options.
- Third? Well we already have B, C or D in our second position, so only two of them could be in the third. Third position is two possibilities.
- The fourth is only one.
- The fifth is only one because either A or E is already in the first position.
Thus, we are left with 2 x 3 x 2 x 1 x 1 = 12 possibilities. Desired outcomes over total outcomes = 4/12 =1/3.
With this problem, we could have listed out all the possibilities and calculated desired and total from that, but this method can be used in many different problems and saves a lot of times. Happy studying!
David Greenslade is a Veritas Prep SAT instructor based in New York. His passion for education began while tutoring students in underrepresented areas during his time at the University of North Carolina. After receiving a degree in Biology, he studied language in China and then moved to New York where he teaches SAT prep and participates in improv comedy.