Last Updated On: July 1st, 2024
The Math section of the SAT contains problems pertaining to topics known as combinations and permutations. You may be asked the number of combinations possible in a given situation. For example, let’s say you walk into a restaurant that serves ice cream. You have a choice of five different flavors: chocolate, vanilla, strawberry, mint, and blueberry. You also have a choice of three different toppings: hot fudge, caramel, and strawberry. How many different ways can you choose one scoop of ice cream with one topping?
One way to solve this question (although there are faster ways) is to write the list of all the possible options, systematically. Thus, we can choose chocolate with hot fudge, chocolate with caramel, and chocolate with strawberry. We’ve now exhausted all the options for chocolate ice cream, so now we move to vanilla. We now have vanilla with hot fudge, vanilla with caramel, and vanilla with strawberry. We’re now finished with all the options involving vanilla. Continuing through the list in order, we come up with the following:
strawberry with hot fudge
strawberry with caramel
strawberry with strawberry
mint with hot fudge
mint with caramel
mint with strawberry
blueberry with hot fudge
blueberry with caramel
blueberry with strawberry
If you count all the possible options, you’ll see that there are 15 combinations.
Here’s another way to think about this problem: each of the five flavors had three different options for toppings. Five threes means “five times three,” which is 15. So another way to solve combinations problems is to multiply the number of options in each category by each other. Use this method instead of listing all the possible options. It’s faster.
Multiplying works no matter how many categories there are. For example, if you have three different hats, four different shirts, and five different pairs of pants, how many different outfits can you make? The process is easy: 3 X 4 X 5, which equals 60.
Permutations are similar to combinations. Permutations deal with the number of ways that things (or people) can be arranged. For example: let’s say you have five different soccer trophies, and you want to arrange them in a row on your desk. What are all the possibilities of doing so?
Pretend that there are five different blank spaces on a desk. Imagine the first blank space—let’s say it’s the space that’s furthermost left. How many different trophies can you put there? Five. Now let’s pretend you select a trophy and put it in the space. How many options do you have to put a trophy in the blank space next to the first space? Four—because that’s how many trophies you have remaining. Pretend to select a trophy again and put it in that second blank space. How many options do you have to put a trophy in the next blank space? You have three trophies left. Pretend to select a trophy again and put it in that third blank space. Now how many options do you have to put a trophy in the next blank space? Two. Pretend to select a trophy again and put it in that fourth blank space. You have one blank space left. Now how many options do you have to put a trophy in the last blank space? Only one. Now, multiply all those numbers. Why? Because this is a special type of combinations problem; therefore, we follow the abovementioned rule. So 5 X 4 X 3 X 2 X 1 = 120. There are 120 ways that you can arrange five trophies in a row.
You can perform a problem like this on your calculator by typing all those numbers, but there’s an even easier way. If your calculator sports a button with an exclamation mark, you may use that key. That key is called a factorial. Type “5,” hit the factorial key, then press the enter key. You should get the same answer, because 5! means that you multiply all the integers starting with 5 and (always, by definition) ending with 1. (Another example: 7! = 7 X 6 X 5 X 4 X 3 X 2 X 1.)
These simple rules will help you solve the most straightforward math problems involving combinations and permutations.
