Permutations and combinations are counting tools. They have vast applications in probability, especially in determining the number of successful outcomes and the number of total outcomes in a given scenario. Questions about permutations and combinations on the Math IIC will not be complex, nor will they require advanced math. But you will need to understand how they work and how to work with them. Important to both of these undertakings is a familiarity with factorials.

The factorial of a number, n!, is the product of the natural numbers up to and including n:

If you are ever asked to find the number of ways that the n elements of a group can be ordered, you simply need to calculate n!. For example, if you are asked how many different ways 6 people can sit at a table with six chairs, you could either list all of the possible seating arrangements or just answer 6! = 6 5 4 3 2 1 = 720.

A permutation is an ordering of elements. For example, say you’re running for student council. There are four different offices to be filled—president, vice president, secretary, and treasurer—and there are four candidates running. Assuming the candidates don’t care which office they’re elected to, how many different ways can the student council be composed?

The answer is 4! because there are 4 students running for office and thus 4 elements in the set.

Say that due to budgetary costs, there are now only the three offices of president, vice president, and treasurer to be filled, but now six candidates are running. To handle this situation, we will now have to slightly change our method of calculating the number of permutations.

In general, the permutation, _nP_r, is the number of subgroups of size r that can be taken from a set with n elements:

For our example, we need to find ₆P₃:

Consider the following problem:

This problem is a permutation, since the question asks us to order the top three finishers among 10 contestants in a dog show. There is more than one way that the same three dogs could get first place, second place, and third place, and each arrangement is a different outcome. So, the answer is ₁₀P₃ = ^10!/_{(10 – 3)!} = ^10!/_7! = 720.

Graphing calculators and most scientific calculators have a permutation function, labeled _nP_r. In most cases, you must enter n, then press the button for permutation, and then enter r. This will calculate a permutation for you, but often if n is a large number, the calculator can’t calculate n!. If this happens to you, don’t give up! In cases like this, your knowledge of how the permutation functions will save you. Since you know that ₁₀₀P₃ is ^100!/_(100-3)!, you can simplify it to ^100! /_97!, or 100 99 98 = 970,200.

A combination is an unordered grouping of a set. An example of a scenario in which order doesn’t matter is a hand of cards: a king, ace, and five is the same as an ace, five, and king.

Combinations are represented as _nC_r, or , where unordered subgroups of size r are selected from a set of size n. Because the order of the elements in a given subgroup doesn’t matter, this means that will be less than _nP_r. Any one combination can be turned into more than one permutation. _nC_r is calculated as follows:

Here’s an example:

In this example, the order in which the leaders are assigned to positions doesn’t matter—the leaders aren’t distinguished from each other in any way, as they were in the student council example. This distinction means that the question can be answered with a combination rather than a permutation. We are looking for how many different groups of three can be taken from a group of 10:

There are only 120 different ways to elect 3 leaders, as opposed to 720 ways when their roles were differentiated.

There should be a combination function on your graphing or scientific calculator labeled _nC_r. Use it the same way as you use the permutation key.