Knowing this rule–or principle as it is called–will make certain questions on the SAT much, much easier. Before I tell you the rule though, let’s take a look at the following scenario.

*Jane has white socks, red socks, and black socks (she must be a baseball fan!). Additionally, she has four different shoes: tennis, basketball, casual, and golf shoes. If she can wear any pair of socks with any pair of shoes how many different ways can she dress.*

The big question here is should you add the number of shoes and socks or multiply them. One good way to answer this is by doing a little thought experiment (something you can repeat if the solution slips your mind).

Let’s say Jane has got her white socks on. How many different shoe choices that she have? Four, since there are that many pairs of shoes. Now, how many different shoes can she choose from if she has red socks on? The answer, again, is four. At this point, she is already at 8 different shoe-sock getups. So you know that when you are seeing how many ways you can combine two different sets of things, always multiply. Another way of looking at it is for each pair of socks (there are 3 total), Jane has four choices: 3 x 4 = 12.

And that brings us to the Fundamental Counting Principle:

If option #1 has P alternatives and option #2 has Q alternatives (assuming that the two sets of alternatives have no overlap), then total number of different pairs we can form is P*Q.

I know that sounds scary. So don’t labor to wrap your head around it–just remember the logic from Jane and her socks.

## Easier

Hailey, a dog-walker, has to look after three poodles and two dachshunds. There is one bowl for the poodles, since they are quite picky when it comes to cleanliness, and another for the dachshunds, who aren’t nearly as finicky. If a feeding consists of one dachshund and one poodle eating at the same time, how many different ways can Hailey pair off the dogs for feedings?

- 3
- 5
- 6
- 9
- 12

## Medium

A youth symphony orchestra is to consist of five violinists, three clarinetists, and two bassoonists. The conductor wants to put together a trio of each instrumentalist. How many different possible trios are there?

- 5
- 8
- 15
- 24
- 30

## Challenging

A rectangle is broken up into 4 squares. A modern artist can use any of 6 colors to paint the four squares. If no square can be the same color as an adjacent square, and the extreme left square and the extreme right squares are the same colors, how many unique ways can the artist paint the square?

- 120
- 150
- 360
- 750
- 900

## Explanations:

1. There is a lot of verbiage in this problem. All the question is really asking is how many ways can you pair two dachshunds with three poodles: 2 x 3 = 6. (C).

2. We are seeing how many ways we can combine three different sets, so we just multiply the three groups together: 2 x 3 x 5 = 30. Answer (E).

3. Tricky question. The first square can be painted any color. The square next to it can be painted any color, except for the color of the first square. So we have 5 choices for the second square. For the third square, it can be any color that is not the same color of the second square or the same color as the first/last squares (which are the same color). So we have 4 options for the third square.

And for the last square? Well, the twist is that it must be the same color as the first square, so we actually don’t have a choice. In the end, we are left with 6 possibilities for the first square, 5 for the second square, and 4 for the third square: 6 x 5 x 4 = 120. Answer (A).

Isn’t Q2, answer E ? D is 10. 2*3*5=30 which is option E

Yes, that was a small typo 🙂

It should read (E).

I have a question about the challenge question. The 6 choices for the first and 5 for the second squares are ok, but the third square it seems to me has only 4 possible choices as it cannot be either the same as the first (which will be same as last which will be adjacent to the third) or the second, which will be adjacent to the third. 6x5x4 would give 120 possibilities.

Hi Lisa,

You’re right about this! Great catch. I’ve updated the post 🙂