# SAT Math: Listing and Counting

Problems that ask you to determine in how many ways something can happen are generally called permutation and combination problems.  They are not as common on the SAT as, say, function problems, but if you are looking to maximize your math score, it would be helpful to master these types of problems.  You are likely to run into one or two of them, and when you do they are usually near the end of a section among the harder problems.

Permutations are problems in which order matters.  For instance, if you are asked to determine in how many ways a group of students can line up, that is a permutation problem.  If order does not matter, then using combinations is appropriate.  You might recall the formulas from your math class:

Some problems on the SAT can be solved directly using one of these formulas.  Consider this problem for instance:

A chord is a line segment that connects two points on a circle as shown below:

If you choose 7 distinct points on the circle, how many different chords can be drawn connecting any two of these points?

(A)  7
(B)  13
(C)  21
(D)  42
(E)  64

This is a combination problem because the order that you select the points does not matter (connecting Point A to Point B gets you the same chord as connecting Point B to Point A). You can use the combination formula to solve this problem like this:

$_{7}C_{2}=\frac{7!}{2!\cdot (7-2)!}=\frac{7!}{2!\cdot 5!}=\frac{7\cdot 6}{2\cdot 1}=\frac{42}{2}=21$

The correct answer is Choice (C).

You may have noticed, however, that the title of this post is “Listing and Counting,” not “Permutations and Combinations.”  Many of the problems of this type that appear on the SAT can’t be solved with a  nice, neat permutation or combination calculation.  Let’s look at a couple of ways that we can approach these problems.

Listing
You can probably guess what this technique is all about.  In order to figure out in how many ways something can happen, you make a list of all the possibilities.  We actually could have done this for the “chord” problem above.  Imagine that we call the seven points A, B, C, D, E, F and G.  We can make an organized list of all the possible chords connecting two of these seven points as follows:

AB  AC  AD  AE  AF  AG
BC  BD  BE  BF  BG
CD  CE  CF  CG
DE  DF  DG
EF  EG
FG

A quick count tells us that there are 21 possible chords, the same answer we got by employing the combination formula.

Here’s another (slightly harder) problem where listing can help us figure out the correct solution:

There are four seniors and five juniors who are members of the Community Service Committee at the local Boys & Girls Club.  A group of three of them will be selected to go to a national conference.  If the group must contain two seniors and one junior, how many possible groups of three students could be made to attend the conference?

(A)  15
(B)  30
(C)  120
(D)  360
(E)  504

Let’s list out the different ways that we can pair up the four seniors into groups of two.

S1S2      S1S3      S1S4
S2S3       S2S4
S3S4

There are six possible combinations of two seniors.  Each of the five juniors can be paired up with each of these pairs of seniors, so we can have 6 x 5 = 30 different groups of students who could attend the conference.  The correct answer is Choice (B).

Counting
This technique allows us to use the Counting Principle to get our solution, sometimes in clever ways.  Consider this problem:

Sandra is planning a trip in which she will visit the following five cities:  Albuquerque  Boston, Chicago, Dallas and Evanston.  In how many different ways can she arrange the order that she visits the cities if her trip must end in either Chicago or Evanston?

(A)  6
(B)  12
(C)  24
(D)  48
(E)  120

We can start by laying out blanks to fill with our numbers (sort of like hangman).

_____    _____    ______    ______    ______

Since we know that she needs to end her trip in Chicago or Evanston, we want to start by filling in that last blank first.  Because there are only two possible cities that can end her trip, we’ll fill in a two in that blank.

______    _____    ______    ______    __ 2___

Whatever city ends up being the last she visits, can’t be the first she visits, so there are only four possible cities for that first blank.  Similarly, the second city can’t be the one she visits first or the one she visits last, so there are only three possibilities for that blank, and so on.

__4___    __3__    __2___    __1___    __ 2___

The total number of ways that Sandra could arrange her trip is 4 x 3 x 2 x 1 x 2 = 48.  The correct answer is Choice (D).

Here’s a second problem where this technique would be useful:

How many three digit numbers are there in which all three digits are unique (in other words, no digit repeats)?  For example, 323 would NOT be such a number because the 3 is repeated.

(A)  504
(B)  612
(C)  648
(D)  720
(E)  900

Start by laying out your three blanks like so:

______    ______  ______

There are ten digits all together (0 – 9), but a three-digit number can’t start with zero, so we have only nine possibilities for that first digit.

___9___    ______    ________

For the second digit, we can use any digit from 0 – 9, except the one we used as the first digit. Therefore, there are nine possibilities for the second digit as well.

___9___    ___9___    ________

Finally, the third digit can be any number from zero through nine except for the two numbers that were used for the first two digits, leaving us with eight possibilities.

___9___    ___9___    ____8___

There are 9 x 9 x 8 = 648 three-digit numbers in which all of the digits are unique.  The correct answer is Choice (C).

For more practice on “Listing and Counting” problems, go here.