Permutation
In short: A permutation is an ordered arrangement of items chosen from a larger group, and the number of ways to arrange r out of n different items is nPr = n!/(n – r)!.
PrecalculusProbabilityS-CP.B.9
What is Permutation?
A permutation is a selection in which order matters. Choosing a captain, a programmer and a driver from a robotics club is a permutation, because swapping two members’ jobs produces a different team plan.
The number of ways to arrange \(r\) of \(n\) distinct items is \(P(n, r) = \frac{n!}{(n – r)!}\). Equivalently, multiply \(r\) descending numbers starting at \(n\). When some items are identical, divide by the factorial of each repeat count: the letters of a word with repeats can be arranged in \(\frac{n!}{p!\,q!\cdots}\) ways.
How to spot it
- A problem assigns different titles, ranks or positions, such as first, second and third place.
- The words arrangement, order, schedule or line up appear.
- Swapping two chosen items would count as a new outcome, as with a locker code or a seating chart.
- A word or number string contains repeated symbols and the question asks for distinct orderings.
The method
- Decide whether order matters; if it does, this is a permutation.
- Count the choices for each position in turn, reducing by one each time when items cannot repeat.
- Multiply the counts, or use \(\frac{n!}{(n – r)!}\).
- If items repeat, divide by the factorial of each repeat count to avoid counting identical swaps.
Permutation: solved example
Problem
A robotics club in Ohio has 11 members and must name a captain, a programmer and a driver, all different people. How many role assignments are possible? Also, how many distinct arrangements of the letters in BALLOON are there?
Solution
Order matters: \(P(11, 3) = 11\cdot 10\cdot 9 = 990\) assignments.
BALLOON has 7 letters, with L twice and O twice: \(\frac{7!}{2!\,2!} = \frac{5040}{4} = 1260\). The answers are 990 and 1,260.
Try it
Question 1
Compute \(P(6, 2)\).
Show the answer
\(6\cdot 5 = 30\).
Question 2
How many distinct arrangements are there of the letters in PEPPER?
Show the answer
6 letters with P three times and E twice: \(\frac{6!}{3!\,2!} = \frac{720}{12} = 60\).
Permutation vs. Combination
With a permutation, the same people in different roles count as different outcomes. With a combination, only the group matters, so each group is counted once. For the same n and r, the ordered count is always larger, by the factor \(r!\).
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