Additional Mathematics notes, chapter 4
Permutation and Combination
Textbook Additional Mathematics Form 5 (DLP), pages 118–139
What this chapter covers
- Use the multiplication (product) rule to count the number of ways to complete a task made of several steps (4.1.1)
- Use factorial notation and find the number of permutations of n different objects, r objects at a time, and in a circle (4.1.2)
- Find permutations of objects with identical items and permutations that meet conditions, and solve problems (4.1.2, 4.1.3)
- Explain the difference between a permutation and a combination, and find the number of combinations nCr (4.2.1, 4.2.2)
- Solve problems that involve combinations with conditions (4.2.3)
4.1Permutation
The multiplication rule
Multiplication rule (product rule). If one task is done in m ways and, for each of those, a second task is done in n ways, then the two tasks together can be done in m × n ways. The rule extends to any number of steps.
| Situation | Rule |
|---|---|
| Steps happen one after another ("and") | Multiply the numbers of choices |
| Only one of several separate options ("or") | Add the numbers of ways |
Example (textbook p. 121): choosing a set meal
A cafe offers 3 main dishes and 4 drinks. A set is one main dish and one drink.Number of sets = 3 × 4 = 12.
A tree diagram lists every possible outcome. Use it to see why the rule works, then use multiplication for bigger problems.
Factorials and permutations of n objects
Factorial. n! = n × (n − 1) × (n − 2) × … × 2 × 1, with 0! = 1. The number of permutations (arrangements in a row) of n different objects is n!.
Order matters in a permutation: AB and BA are different arrangements. Reason: the first place can be filled in n ways, the second in (n − 1) ways, and so on.
Example (textbook p. 122): arranging letters
Letters of the word MATHS: 5 different letters.Arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120.
Simplify a quotient of factorials by cancelling, not by calculating both: 9!/7! = 9 × 8 = 72. Do not write 9!/7! = 2!.
Permutations of r objects from n different objects
nPr = n! / (n − r)! is the number of arrangements of r objects chosen from n different objects (r ≤ n).
Example (textbook p. 124): a leader and an assistant
Choose a class leader and an assistant from 8 pupils. The two posts are different, so order matters.8P2 = 8 × 7 = 56.
Quick check. Calculate 7P3.
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7P3 = 7 × 6 × 5 = 210.
Circular permutations
Circular permutation. n different objects arranged in a circle can be arranged in (n − 1)! ways. Rotating the whole circle does not give a new arrangement, so fix one object and arrange the other (n − 1).
Example (textbook p. 123): round table
6 people sit at a round table: (6 − 1)! = 5! = 120 ways.If two particular people must sit together: treat them as one block, 2! × (5 − 1)! = 2 × 24 = 48.
If clockwise and anticlockwise orders look the same (a bracelet, a necklace), divide by 2: (n − 1)!/2.
Permutations with identical objects
For n objects where a are identical of one kind, b identical of another and c of a third, the number of permutations is n! / (a! b! c!).
Example (textbook p. 127): letters of a word
BANANA has n = 6 letters: A three times, N twice.6! / (3! 2!) = 720 / 12 = 60 arrangements.
Permutations with conditions
| Condition | Method |
|---|---|
| Some objects must be together | Treat them as one block; arrange the block with the rest, then arrange inside the block |
| Some objects must not be together | Total arrangements − arrangements with them together |
| A place has a restriction (first digit, last digit, even number) | Fill the restricted place first, then the others |
| Objects of one type must be separated | Arrange the other objects first, then put the type in the gaps |
Example (textbook p. 128): books on a shelf
5 different books, 2 particular books together: block + 3 others = 4 objects, 4! = 24; the 2 books swap, 2! = 2. Total 24 × 2 = 48.Not together: 5! − 48 = 120 − 48 = 72.
Example (textbook p. 129): even numbers
Even 4-digit numbers from the digits 1, 2, 3, 4, 5, 6 with no repeated digit: last digit 2, 4 or 6 (3 ways), then the other three places from the remaining 5 digits: 5P3 = 60. Total 3 × 60 = 180.A number cannot start with 0. When 0 is available, deal with the first place (and the last place for even numbers) before the others.
Quick check. How many ways can the letters of SCHOOL be arranged?
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6 letters, O twice: 6!/2! = 360.
4.2Combination
Permutation or combination?
| Permutation | Combination | |
|---|---|---|
| Order | Matters | Does not matter |
| Meaning | Arrange | Select |
| Example | President and secretary from 7 people | Two representatives from 7 people |
nCr = n! / (r! (n − r)!) is the number of ways to select r objects from n different objects when order is not important. Notice nPr = r! × nCr.
Example (textbook p. 133): a committee
Select 3 members from 8 people: 8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 56.The same people selected in a different order are the same committee.
nCr = nC(n − r). Choosing which 2 of 9 to leave out is the same as choosing which 7 to take: 9C2 = 9C7 = 36. Also nC0 = nCn = 1.
Combinations with conditions
- Particular objects included: choose them first, then select the rest from what is left.
- Particular objects excluded: remove them, then select from the rest.
- Two groups: select from each group and multiply (and), for example 5C2 × 4C2.
- At least one: total − none, or add the cases (or).
Example (textbook p. 135): boys and girls
A team of 4 is chosen from 5 boys and 4 girls.No condition: 9C4 = 126.
2 boys and 2 girls: 5C2 × 4C2 = 10 × 6 = 60.
At least 1 girl: 126 − 5C4 = 126 − 5 = 121.
Do not use 5C2 + 4C2 when both groups are needed. "And" means multiply; add only for separate cases such as "2 boys or 3 boys".
Quick check. Find 10C3 and say whether choosing a gold, silver and bronze medallist from 10 runners uses it.
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10C3 = 120. The medals are different, so the runners are arranged: 10P3 = 720, not 10C3.