What are permutations and combinations? In mathematics, they are two different things, and have specific definitions. However, you do not need to know the definitions for the SAT (if you really want to know, look at Tip #2). In everyday English, the two terms are often used interchangeably. They refer to the number of ways of doing something.

For example, there are 2 possible outcomes when you flip a coin (assuming it can’t land on its edge)

If you roll a die, there are 6 possible outcomes.

So don’t worry about the terms “permutations” and “combinations.” You can just say “outcomes,” “orderings,” or “arrangements” (which is the term you’ll usually see on the SAT).

What is probability? Well, it’s the likelihood that something will occur. In everyday life, probability is usually expressed as a percentage, and it’s often very difficult to determine.

What is the probability that our lacrosse team will make the playoffs?

What is the probability that President Obama will be re-elected?

In math (including the SAT), it’s a different matter. Probabilities are expressed on a scale of 0 to 1 (which is equivalent to 100%), and they can often be determined exactly. Each probability is a fraction equal to (number of arrangements that qualify)/(total number of possible arrangements).

Q. If a die is rolled, what is the probability that an even number will appear on top?

There are 6 possible outcomes {1,2,3,4,5,6}, and 3 of them are even, so the answer is 3/6 or 0.5.

Of course, you won’t usually encounter such an easy question on the SAT. So here are some tips to help:

**Tip #1 (basic) Multiplication Rulez!**

Sorry for the spelling – even nerds want to sound cool once in a while.

This tip is mostly self explanatory – multiply everything (don’t add).

Q. If a fair coin is flipped twice, what is the probability of seeing heads both times?

0.5 X 0.5 = 0.25

In other words, the probability of two events occurring in succession equals the product of the individual probabilities (P[event 1] X P[event 2]).

Q. If two dice are rolled separately, what is the probability of rolling a 3 followed by an even number?

1/6 X 3/6 = 1/12 or 0.83

Q. If a random number from 1 to 10 is chosen, and a random letter from A to Z is chosen next, how many different arrangements are possible?

10 X 26 = 260

Q. A company that has 5 salespersons wants to send one of them to New York and the other to Minnesota. How many different ways are there to choose them?

Say you pick the New York salesperson first – you have 5 to choose from. Then you’ll have 4 left over to choose for the Minnesota job. That’s 5 X 4 = 20 possibilities. Note that it doesn’t matter which state you pick first.

**Tip #2 (intermediate) 2 Important Questions**

On harder questions, you need to consider a couple of circumstances that affect the number of orderings.

**Question A: Does order matter?**

Consider the following 2 questions:

Q. A cheerleading squad has 8 members. If 2 co-captains are to be chosen, how many different arrangements are possible?

Q. A booster club has 8 members. If a President and a Vice President are to be chosen, how many different arrangements are possible?

At first glance, you might think that both questions have the same answer. But hold on – if you choose Arlene as the first cheerleader and Pablo as the second, isn’t that exactly the same as choosing Pablo first and Arlene second? But that doesn’t hold true for the booster club. Ricky for president and Jennifer for VP certainly isn’t the same as Jennifer for President and Ricky for VP.

In the first case (cheerleaders), order doesn’t matter. For every ordering X, then Y, there will be a reverse ordering that you shouldn’t count. So you need to divide your answer by 2 (there are (8 X 7)/2 = 28 arrangements).

In the second case, order does matter, so you should count every arrangement (8 X 7 = 56 possible).

Most of the time, order will matter, and you don’t have to worry. But if you’re picking several interchangeable things from a group, you need that extra step.

Every time I’ve seen this come up on the SAT (order doesn’t matter), you are asked to choose 2 or 3 things from a group. If you have to choose 2 things, you know you need to divide the number of arrangements by 2 to get the number of unique orderings. What if you’re asked to choose 3 things from a group? Then you need to divide by 6! Why 6? Well, suppose you choose A, B, and C. There are 3 X 2 X 1 = 6 possible ways to order them (ABC, ACB, BAC, BCA, CAB, CBA). And there will be six ways to order any subgroup (such as ADE, etc.), and you only want to count each subgroup once.

Just in case you ever have to choose 4 things from a group, remember to divide by 24 (4! = 4 x 3 X 2 X 1).

By the way, this is the difference between the mathematical terms “permutation” and “combination.” If order matters, it’s a permutation.

**Question B: Is there replacement?**

Q. If a bag contains 4 blue marbles and 4 red marbles, and 2 marbles are taken, what is the probability that they will both be blue?

The wording must be changed to specify whether or not you can take one marble, *put it back and the bag*, and then choose another.

No replacement: 4/8 X 3/7 = 12/56 = 3/14

Replacement (first marble replaced before second is chosen): 4/8 X 4/8 = 16/64 = 1/4

Replacement is sometimes called *repetition*.

Q. A butterfly collecting club has 8 members. If a President and a Treasurer are to be chosen, how many different arrangements are possible?

Can the same person serve as both president and treasurer? If you don’t know, you can’t answer the question.

If two different people must serve, there are 8 X 7 = 56 arrangements. If someone can hold both offices (i.e. repetition is possible), it’s 8 X 8 = 64.

Usually, there won’t be replacement on the SAT. Just watch out for it.

**Tip #3 (advanced) Do Restricted Things First**

On all of the questions we’ve looked at so far, the situation involved unrestricted events. You could pick any member as president, or any marble first, etc. But on harder SAT questions, you will be presented with restricted choices.

Q. Ahmad, Barbara, Chen, DeShawn, and Enrique will each read a poem in English class during the coming week. One student will read a poem on each day of the week, and only Ahmad or Chen will be permitted to read on Wednesday. How many arrangements are possible?

The trick here is to choose Wednesday’s speaker first, since Wednesday is restricted. You can choose either of 2 students (A or C) on that day. Now you can choose in any order you like, since the other days aren’t restricted. If you go from early to late, you’ll have a choice of 4 students for Monday (since one was already chosen), 3 for Tuesday, 2 for Thursday, and 1 for Friday. So the answer is 2 X 4 X 3 X 2 X 1 = 48.

Q. Ahmad, Barbara, Chen, DeShawn, and Enrique will each read a poem in English class during the coming week. One student will read a poem on each day of the week, and Ahmad must read on Tuesday, Thursday, or Friday. How many arrangements are possible?

Now Ahmad is restricted, instead of Wednesday. So we do a person first instead of a day. There are 3 possible days for Ahmad, and then there will be 4 days left to choose for Barbara, etc. 3 X 4 X 3 X 2 X 1 = 72.

Finally, geometric probability questions are quite easy. If an object is tossed into Area A, what is the probability that it will land in Area B (a smaller area within Area A)? Answer: Area B/Area A.

If you study hard, there is a high probability that you’ll do well on your test!