# Tackling Rare SAT Combinations and Permutations

Combinations and permutations are not common on the SAT. They normally appear on the last page of the math section and are almost always presented as a word problem. If you want to achieve a score over 700 on the math section of the SAT, you will need to practice these, as they involve choosing for a group (combination) or arranging a group (permutation). While arranging a group (permutations) can seem rather simple, arranging the order of things can actually be somewhat difficult. In fact, students frequently overlook the aspect of permutation and combination. For instance, “repetition” means the same choice can be used twice or even more and students may not realize this. Here is an example of a permutation question with repetition:

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

This probably seems like a nightmare of a question, but what we do know is that there are 10 possible values starting with 0 and ending at 9. Since there are four digits, the total possibilities are 10 x 10 x 10 x 10 x 10 = 10,000. Repetition is allowed because the code could be 555 or 999, for instance.

Then here is another example:

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

This question doesn’t allow for repetition because each digit has to be unique, which means it has to be different from the others. So for the first digit, there will be 10 choices, for the second there are 9 choices, for the third there are only eight options, and so on. You will multiply like this: 10 x 9 x 8 x 7 = 5,040 combinations.

Questions, such as how many ways seven cats and dogs can be placed first and second in line are permutations and not combinations. To solve this problem, you will take the 7 cats and dogs and multiply like this: 7 x 6 x 5 x 4 x 3 x 2 x 1 and divide it by 5 because that is the number of remaining cats and dogs after two have been put in their places. This means the problem will look like this: 7 x 6 x 5 x 4 x 3 x 2 x 1/5 x 4 x 3 x 2 x 1. You will cancel out 5 x 4 x 3 x 2 x 1 on both sides, leaving 7 x 6, which means there are 42 ways. This problem is not a combination because order matters and the first and second positions are different. If order didn’t matter and the first and second positions were the same, it would be a combination.

To succeed on these problems, you simply look at the wording so you know what kind of math you are doing.

Image source: Toni Blay