Counting & exam skills

Permutation & combination: does order matter?

Build counts from choices, distinguish arrangements from selections, and avoid counting the same outcome more than once.

AdvancedAbout 16 min3 worked examples
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01 / THE STARTING POINT

Before you begin

What you need: Whole-number multiplication and factorial notation.

  • Use the multiplication principle
  • Distinguish permutations from combinations
  • Handle simple restrictions and repeated objects
02 / BUILD THE CONCEPT

Understand the idea

Count choices in stages

If a task has a choices at the first stage and b choices at the second for every first-stage choice, there are ab outcomes. For example, 3 shirts and 2 trousers make 6 outfits.

Arrangements care about order

Selecting a captain then a vice-captain creates distinct ordered roles. From n distinct people without repetition, the choices are n, then n − 1, and so on. Their product is a permutation.

Selections ignore order

A committee does not distinguish selection order. Each group of r distinct people appears r! times among ordered arrangements, so divide by r!. Factorial n! multiplies integers from 1 to n; 0! = 1.

03 / A DIFFERENT WAY TO SEE IT

Choosing two from A, B and C

SelectionIts ordered arrangements
A and BAB, BA
A and CAC, CA
B and CBC, CB
Three unordered pairs produce six ordered arrangements: each pair is counted 2! times.
04 / FROM IDEA TO ANSWER

A method you can reuse

KEEP THIS HANDY

nPr = n!/(n − r)!; nCr = n!/[r!(n − r)!]

For integers 0 ≤ r ≤ n, distinct objects, and selection without repetition.
  1. Decide whether order or assigned roles make outcomes different.
  2. Check whether repetition is allowed and whether objects are distinct.
  3. Count the valid stages, then divide out duplicates only when justified.
A useful insight

nCr = nC(n − r): choosing who joins a group determines exactly who stays out.

05 / WATCH THE METHOD WORK

Worked examples

Read the question first. Try a step yourself, then compare your reasoning.

EXAMPLE 01

Arrange books

How many ways can 4 distinct books be arranged in a row?

  1. There are 4 choices for the first position.
  2. Then 3, 2 and 1 choices remain.
  3. Multiply 4 × 3 × 2 × 1.
Answer24 arrangements
EXAMPLE 02

Choose a committee

Choose 2 people from 5 for a committee with no distinct roles.

  1. Order does not matter.
  2. 5C2 = (5 × 4)/(2 × 1).
  3. Divide by 2! because each pair is otherwise counted twice.
Answer10 committees
EXAMPLE 03

Repeated letters

How many distinct arrangements of the letters in MOM?

  1. Treating all three positions as distinct gives 3! = 6.
  2. The two M letters are identical; swapping them changes nothing.
  3. Divide by 2!.
Answer3 arrangements: MMO, MOM, OMM
06 / YOUR TURN

Check your understanding

Quick questions, with explanations. These are for self-study and do not affect your account score.

0 of 3 questions checked

QUESTION 1Choose a captain and vice-captain from 4 people. Ways?

See the explanation

12. The roles are different: 4 × 3 = 12.

QUESTION 2Choose all 5 people from a group of 5. Ways?

See the explanation

1. There is exactly one group containing everyone: 5C5 = 1.

QUESTION 3A 3-digit code permits leading zero and repeated digits 0–9. How many codes?

See the explanation

1000. Each of the three slots has 10 choices: 10³ = 1000. This is a code, not a three-digit integer.

07 / TAKE THE NEXT STEP

Put your understanding to work

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