Permutation vs. Combination

A permutation and a combination both count the number of ways to pick items from a group, but they answer slightly different questions. A permutation counts the number of ways to arrange or order a set of items, where the order matters. A combination counts the number of ways to select items from a group, where the order does not matter. That single distinction, whether order counts, is the entire difference between the two ideas, and it determines which formula you should reach for.

In a permutation, swapping the order of the same items produces a different outcome. Think of a race: first place, second place, and third place are three different medals, so the arrangement (Amy, Ben, Cara) is not the same result as (Ben, Amy, Cara), even though the same three people are involved.

The number of permutations of \(n\) items taken \(r\) at a time is written \(P(n, r)\) or \(_nP_r\), and is calculated as:

\( _nP_r = \dfrac{n!}{(n-r)!} \)

Here \(n!\) means \(n\) factorial, the product of all positive integers up to \(n\). If you need a refresher on factorials, including why \(0! = 1\), it's worth reviewing factorial notation before working through permutation problems.

In a combination, the order of selection is irrelevant, only which items end up chosen matters. Think of picking a committee of 3 people from a group of 10: choosing (Amy, Ben, Cara) is exactly the same committee as choosing (Cara, Amy, Ben). There is no "first pick" that outranks the others.

The number of combinations of \(n\) items taken \(r\) at a time is written \(C(n, r)\) or \(_nC_r\) (also read as "\(n\) choose \(r\)"), and is calculated as:

\( _nC_r = \dfrac{n!}{r!(n-r)!} \)

Notice that the combination formula is just the permutation formula divided by \(r!\). That division removes all the duplicate orderings of the same \(r\) items, since a combination doesn't care how those \(r\) items were arranged.

Before reaching for a formula, ask yourself one question: if I rearranged the same chosen items, would that count as a new, different result?

• If yes, order matters, use a permutation. Examples: assigning gold, silver, and bronze medals; creating a PIN code from digits; ranking finalists in a contest; arranging books on a shelf.
• If no, order does not matter, use a combination. Examples: choosing a committee; picking lottery numbers; selecting toppings for a pizza; forming a study group from a class.

A useful mental shortcut: if the word "arrange," "order," "sequence," or "rank" fits naturally in the problem, lean toward a permutation. If the word "select," "choose," or "group" fits better, lean toward a combination.

Suppose you have 5 friends and want to know how many ways you can pick 2 of them.

As a combination (just picking who goes, order doesn't matter):

\( _5C_2 = \dfrac{5!}{2!(5-2)!} = \dfrac{5!}{2! \cdot 3!} = \dfrac{120}{2 \cdot 6} = 10 \)

There are 10 possible pairs of friends.

As a permutation (picking a first-place and second-place finisher between them, order matters):

\( _5P_2 = \dfrac{5!}{(5-2)!} = \dfrac{5!}{3!} = \dfrac{120}{6} = 20 \)

There are 20 possible ordered results, exactly double the combinations, because each pair of friends can be arranged in \(2! = 2\) ways (first and second, or vice versa). This is exactly why \( _nP_r = {_nC_r} \times r! \) for any \(n\) and \(r\): a permutation count is always the combination count multiplied by the number of ways to order the chosen items.

Since \( _nC_r = \dfrac{_nP_r}{r!} \) and \(r! \ge 1\), the number of combinations can never exceed the number of permutations for the same \(n\) and \(r\). They are only equal when \(r! = 1\), which happens when \(r = 0\) or \(r = 1\), since choosing zero items or a single item leaves no room for a different order.

Once you're comfortable telling permutations and combinations apart, the next step is combining these counting tools with probability. See probability with permutations and combinations for how to turn these counts into actual probabilities.

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