# SAT Tip: Cracking the Permutation-Question Code

*This tip on improving your SAT score was provided by Vivian Kerr at Veritas Prep.*

Combinations and permutations are fairly rare on the SAT. They usually appear on the last page of a Math section and are almost always presented in some type of word problem. If you’re looking for a 700+ math score, you’ll definitely want to practice them. These types of questions usually involve arranging a group (permutation) or choosing from a group (combination).

Questions that involve arranging a group are “permutations,” since in an arrangement the order of things is important. For those of us bad spellers out there, this especially hits home (i.e., “friend” vs. “freind”). “Repetition” means the same choice can be used more than once. This aspect of combination and permutation questions is often overlooked by students. Let’s first look at a permutation question with repetition:

** A safe uses a three-digit code. How many possible codes are there?**

For every digit we know, we have 10 possible values: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Since we have three digits, the total possibilities will be 10 x 10 x 10 = 1000. In this question, repetition is allowed because the code could be 777 or 888, for example.

But what if the question looked like this one?

** A safe uses a three-digit code where each digit is unique. How many possible codes are there?**

This question does not allow repetition because each digit must be unique, or different. For the first digit, we still have 10 choices. For the second digit, we would have only nine options because we couldn’t re-use the first digit. For the third digit, we would have only eight options. So 10 x 9 x 8 = 720 codes. Try deciphering a question on your own for a moment. Does this question involve repetition?

** How many ways can seven boys and girls line up?**

If you answered, no, you’re correct. This is a simple permutation: 7 x 6 x 5 x 4 x 3 x 2 x 1. When we’re dealing with people, repetition usually does not occur because you can’t have the same person hold two positions physically. Let’s adjust this question slightly, just to practice the permutation formula: n! / (n-k)!. The formula makes use of factorials (!) which are the product of all positive numbers that are less than or equal to n. The factorial of 3 is 3 x 2 x 1, or 6. Now consider the following question:

**How many ways can seven boys and girls be placed first and second in line?**

Here we’re choosing from seven boys and girls, so n = 7, and we have two “slots” to place them in, so k = 2. This is a permutation—not a combination—because first and second are different positions and the order matters. To solve n! / (n-k)! involves the following steps:

7! / (7-2)! =

7! / 5! =

7 x 6 x 5 x 4 x 3 x 2 x 1 / 5 x 4 x 3 x 2 x 1=

7 x 6 ~~x 5 x 4 x 3 x 2 x 1~~ / ~~5 x 4 x 3 x 2 x 1~~=

7 x 6=

42 ways

Always remember to first identify whether repetition is allowed. Then calculate the individual possibilities and multiply those individual possibilities together. Especially be on the lookout for permutation questions involving digits. These are the ones in which knowing the difference between allowing and not allowing for repetition will make the big difference.

*Vivian Kerr has been teaching and tutoring in the Los Angeles area since 2005. She graduated from the University of Southern California, studied abroad in London, and has worked for several test-prep giants tutoring, writing content, and blogging about all things SAT, ACT, GRE, and GMAT.*

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