Grade by Grade
A. Counting and Probability · Permutations and Combinations

Permutations: Ordered Selections of Distinct Objects

A2.2 A2.1About 70 minutes
Learning goals
  • I can explain what a permutation is and recognize when order matters.
  • I can calculate P(n, r) using the formula n!/(n − r)! and the counting principle.
  • I can use standard notation such as P(n, r) and ₙPᵣ.
  • I can solve equations involving P(n, r).

Words to know

permutation
An ordered arrangement of objects. A permutation of r objects chosen from n distinct objects is an ordered selection, so ABC and CBA are different permutations.
P(n, r)
The number of permutations of r objects chosen from n distinct objects: P(n, r) = n!/(n − r)!. Also written ₙPᵣ or nPr.
order matters
Rearranging the chosen objects gives a different outcome, for example first and second place, or the digits of a code.

Gold, silver and bronze

Eight skaters reach the final of a speed-skating race. In how many ways can the gold, silver and bronze medals be awarded?

Three positions, filled in order, with no skater winning two medals: 8 choices for gold, then 7 for silver, then 6 for bronze. 8 × 7 × 6 = 336.

This is a permutation: we chose 3 of the 8 skaters, and the order of the three matters (winning gold is not the same as winning bronze).

The permutation formula

The number of permutations of r objects chosen from n distinct objects is

P(n, r) = n × (n − 1) × (n − 2) × … × (n − r + 1) (r factors)

Multiply by (n − r)!/(n − r)! to write it with factorials:

P(n, r) = n!/(n − r)!

Other notations you will see: ₙPᵣ, nPr (on calculators), and ⁿPᵣ. In the medal question, P(8, 3) = 8!/5! = 8 × 7 × 6 = 336.

Special cases: P(n, n) = n!/0! = n!, which is why 0! = 1 is so useful. P(n, 1) = n, and P(n, 0) = n!/n! = 1 (one way to choose nothing).

10987654321the 4 positions we fill6! that we divide outP(10, 4) = 10! ÷ 6! = 10 × 9 × 8 × 7
P(10, 4) keeps the first 4 factors of 10!. Dividing by 6! removes the 6 factors for the positions we do not fill.
A first:ABACADB first:BABCBDC first:CACBCDD first:DADBDCP(4, 2): ordered pairs from A, B, C, D4 rows × 3 pairs= 4 × 3= 12AB and BA aredifferent here.
All the permutations of 2 letters chosen from A, B, C and D. P(4, 2) = 4!/2! = 12. Swapping the two letters gives a different permutation.
Worked example

Example 1: The executive committee

A student organization has 10 members. It needs an executive committee of a president, a vice-president, a secretary and a treasurer, with no one holding two jobs. How many different executive committees are possible?

  1. Each job is a different position, so a committee where Ana is president and Ben is treasurer differs from one where Ben is president and Ana is treasurer. Order matters: this is a permutation.
  2. n = 10 members, r = 4 positions: P(10, 4) = 10!/6!.
  3. 10!/6! = 10 × 9 × 8 × 7 = 5040.

5040 executive committees. (In lesson 7 you will use this number to count committees where the 4 members have equal standing.)

An executive of 4 from 10 members10president×9vice-president×8secretary×7treasurer= 5040
Each job is a position. Once a member is placed, there is one fewer member for the next job.
Worked example

Example 2: Using the calculator

How many 4-letter "words" (any arrangement of letters, whether or not it is a real word) can be made from the 26 letters of the alphabet if no letter is repeated?

  1. Order matters (ABCD ≠ DCBA) and there is no repetition, so this is P(26, 4).
  2. P(26, 4) = 26 × 25 × 24 × 23.
  3. On a calculator: 26 nPr 4 = 358 800.

358 800 words.

Worked example

Example 3: Solving for n

Solve P(n, 2) = 56.

  1. P(n, 2) = n!/(n − 2)! = n(n − 1).
  2. n(n − 1) = 56, so n² − n − 56 = 0.
  3. (n − 8)(n + 7) = 0, so n = 8 or n = −7. n must be a whole number with n ≥ 2.

n = 8. Check: 8 × 7 = 56 ✓.

Does order matter?

Ask: if I swap two of the chosen objects, do I get a different outcome? Swapping the president and the treasurer gives a different executive, so order matters (a permutation). Swapping two pizza toppings gives the same pizza, so order does not matter (a combination, lesson 7). Clue words for order: first, second, president, ranked, arranged, in a row, code, schedule.

Just for fun

A "combination lock" set to 4-17-32 will not open if you dial 32-17-4. Since order matters, mathematicians point out that it should really be called a permutation lock.

Hands-on activity

Podium finishes with playing cards

You need: the ace, 2, 3, 4 and 5 of one suit from a deck of cards (5 cards), and a table.

  1. The cards are 5 skaters. Mark two spots on the table: 1st and 2nd.
  2. Lay out every possible (1st, 2nd) result, one at a time, and record each as a pair such as (3, 5). Work systematically: all results with the ace first, then the 2 first, and so on.
  3. How many results did you find? Compare with P(5, 2).
  4. Add a 3rd spot. Without listing, predict the number of results with P(5, 3). How does each (1st, 2nd) result split into new results?
  5. Swap two cards in one of your results. Is it the same result or a new one? What does that tell you about order?

What to look for: 20 results for two places, since P(5, 2) = 5 × 4 = 20. With a third place each of those 20 results can be finished in 3 ways, giving P(5, 3) = 60. Swapping cards gives a new result, which is exactly why these are permutations.

Your turn

1

Evaluate P(7, 3).

2

Evaluate P(10, 2).

3

A club of 12 members elects a president, a vice-president and a treasurer. How many outcomes are possible?

4

Ten hockey teams play in a tournament. In how many ways can the gold, silver and bronze medals be awarded?

5

How many 3-letter arrangements can be made from the letters of HOCKEY (no letter used twice)?

6

Solve P(n, 2) = 30.

7

Evaluate P(5, 5) and 5!. Explain why they are equal.

More practice: warm-up

1

Evaluate P(5, 2).

2

Evaluate P(6, 1).

3

Evaluate P(8, 0).

4

Evaluate P(4, 4).

5

Evaluate ₉P₃.

More practice: on your own

1

How many 5-digit codes can be made from the digits 1 to 9 if no digit repeats?

2

A DJ chooses and orders 4 songs from a list of 15. How many playlists are possible?

3

Solve P(n, 3) = 60.

4

Three different prizes are given to 3 of 20 people. How many ways if (a) no one can win more than one prize, (b) a person may win more than one?

5

Show that P(n, r) = n × P(n − 1, r − 1).

Challenge

1

How many arrangements of 1, 2 or 3 letters can be made from the letters of MAPLE, with no letter repeated in an arrangement?

2

Solve P(n, 2) + P(n, 1) = 100.

3

Prove that P(n, r + 1) = (n − r) × P(n, r).

Check your understanding

Quiz

1.

What is P(6, 2)?

2.

Which formula gives P(n, r)?

3.

From 10 members, how many ways can a president, vice-president, secretary and treasurer be chosen?

4.

What is P(8, 8)?

5.

Twelve runners race. In how many ways can 1st, 2nd and 3rd place be decided (no ties)?

6.

Which situation is counted with permutations?

For parents and teachers

Big idea (A2.2, A2.1): a permutation is an ordered selection. P(n, r) = n!/(n − r)! is just the multiplicative counting principle with r shrinking slots: n × (n − 1) × … (r factors).

How to help: For each problem, ask "If you swap two of the chosen items, is it a different outcome?" If yes, order matters and a permutation fits. Encourage your teen to check a calculator nPr answer by writing out the r factors.

Common difficulties:

  • Dividing by r! as well (that gives combinations, lesson 7).
  • Using P(n, r) when repetition is allowed. With repetition, the count is nʳ (see lesson 6).
  • Mixing up n and r: n is the number available, r is the number chosen.
  • Calculator syntax: on most calculators you type n, then nPr, then r.