Permutations and combinations: when order matters

TL;DR:

• A permutation counts ordered arrangements. A combination counts groups where order does not matter.

• Choosing 3 of 5 people gives P(5,3) = 60 lineups but only C(5,3) = 10 groups.

• Each group of r items can be ordered in r! ways, so P = C × r!.

Five friends are trying out for a team, and the coach needs three. One question is "which three make the team?" Another is "who goes first, second and third in the relay?" The first cares only about the group. The second cares about the order. The counts are different, and the gap between them is a neat piece of arithmetic.

The multiplication principle and factorials

If you make one choice in a ways and then a second in b ways, the pair can be made in a × b ways. Lining up 3 people from 5 in order: 5 choices for first place, then 4 for second, then 3 for third. That is 5 × 4 × 3 = 60.

The factorial n! is the product of the whole numbers from n down to 1: 3! = 3 × 2 × 1 = 6, 5! = 120. It counts the ways to arrange n different items in a row.

Worked example: choosing 3 of 5 people

1. Ordered lineups (permutations): P(5,3) = 5 × 4 × 3 = 60. Written with factorials: 5! / (5 − 3)! = 120 / 2 = 60.

2. Each group of 3 people can be arranged in 3! = 6 orders.

3. So the number of groups is the number of lineups divided by the orders per group: 60 / 6 = 10.

4. In symbols: C(5,3) = 5! / (3! × 2!) = 120 / (6 × 2) = 10.

There are 60 ordered lineups but only 10 different groups. Each group was counted 6 times among the lineups, once for each ordering of its three members.

Check by listing a small case

Test the idea on a case small enough to write out. Choose 2 from the four people A, B, C, D. Groups: AB, AC, AD, BC, BD, CD. That is 6, and C(4,2) = 4! / (2! × 2!) = 24 / 4 = 6. Ordered lineups: each group gives 2 orders, so 6 × 2 = 12. And P(4,2) = 4 × 3 = 12. The list matches the formulas.

Deciding which one a problem wants

• Does swapping two chosen items give a different result? A president and a treasurer: yes, order matters. Use permutations.

• Is the result just a set, such as a committee, a hand of cards or a pizza with toppings? Order does not matter. Use combinations.

• A word problem often hints at it: "arrange", "line up", "rank", "code" point to order. "Choose", "select", "group", "team" point to combinations.

A second example: a PIN with no repeated digit

How many 4-digit PINs use four different digits? There are 10 choices for the first digit, 9 for the second, 8 for the third and 7 for the fourth: 10 × 9 × 8 × 7 = 5040. Order matters here, since 1234 and 4321 are different PINs, so this is P(10,4) = 5040. Check with factorials: 10! / 6! = 3,628,800 / 720 = 5040. If repeated digits are allowed, the count is 10⁴ = 10,000.

Compare with choosing 4 digits for a set: C(10,4) = 5040 / 4! = 5040 / 24 = 210. Each set of four digits can form 24 different PINs.

Try three more yourself

• Arranging letters. The word MATH has 4 different letters, so there are 4! = 24 arrangements. For LEVEL, the repeated L and E reduce the count: 5! / (2! × 2!) = 120 / 4 = 30. Divide by the factorial of each repeat, since swapping identical letters changes nothing.

• Pizza toppings. Choose 3 toppings from 8: order does not matter, so C(8,3) = 8 × 7 × 6 / 3! = 336 / 6 = 56.

• Card hands. A 5-card hand from 52 cards: C(52,5) = 52 × 51 × 50 × 49 × 48 / 120 = 2,598,960.

A pattern: C(n, r) = C(n, n − r)

Choosing 3 people to include is the same as choosing 2 to leave out, so C(5,3) = C(5,2). Check: C(5,2) = 5! / (2! × 3!) = 120 / 12 = 10. It matches. This symmetry is a good way to shorten the arithmetic: compute C(10,8) as C(10,2) = 45 instead of using big factorials.

How to check your answer

• Combinations ≤ permutations. C(n,r) can never exceed P(n,r).

• Divide by r!. The ratio between them is r!. Here 60 / 10 = 6 = 3!.

• Small case. Write out all of the possibilities when n is 4 or 5.

• Whole numbers. Counting answers must be whole numbers.

Common mistakes

• Using permutations when order does not matter, which overcounts.

• Forgetting to divide by r!.

• Using n! when only r of the n items are chosen.

• Forgetting that 0! = 1, which is defined that way so that formulas work (C(5,5) = 5!/(5! × 0!) = 1).

• Not noticing that repeated choices are allowed, which changes the count.

Where you will use it next

These counts feed straight into probability. The chance of picking a particular group of 3 from 5 at random is 1/10, which uses the 10 above, as in probability basics with dice. Sloppy conclusions from data are covered in correlation is not causation.

Frequently asked questions

Why is 0! equal to 1?

There is exactly one way to arrange zero items: do nothing. Defining 0! = 1 also keeps formulas like C(n,n) = 1 working.

What is the difference between nCr and C(n,r)?

Nothing. They are different notations for the same number, as are nPr and P(n,r).