Skip to content
IGCSE·Tuition
Additional Mathematics · Lessons

Decide whether order matters

Two counting formulas look almost the same, and choosing the wrong one gives a confident but incorrect answer.

On this page
  1. How do the two formulas differ?
  2. A reliable method
  3. Worked example
  4. The mistake to watch for
  5. Check yourself
  6. Where this leads next

A permutation counts arrangements where order matters, and a combination counts selections where order does not. Ask one question before any calculation: if I swap two of the chosen items, is it a different outcome? This decision appears at the start of every question in permutations and combinations.

How do the two formulas differ?

For n different items, choosing r of them:

  • nPr = n × (n − 1) × … with r factors. For example, 8P3 = 8 × 7 × 6 = 336.
  • nCr = nPr / r!. For example, 8C3 = 336 / 6 = 56.

The division by r! removes the different orders of the same group. Three people can be put in order in 3 × 2 × 1 = 6 ways, so every group has been counted 6 times in the permutation.

A reliable method

  1. Identify what is being chosen (people, digits, letters, cards).
  2. Swap two items in your head. A new outcome means order matters, so use nPr. The same outcome means order does not matter, so use nCr.
  3. Check repeats. The formulas above assume each item is used at most once. If the question allows repeats, use the multiplication principle instead.
  4. Calculate by cancelling, then sense-check that nCr is smaller than nPr.

Worked example

A club has 8 members. Find the number of ways to (a) choose a president, a secretary and a treasurer, and (b) choose a committee of 3.

Part (a): the three roles are different, so swapping Aini and Ben changes who is president. Order matters.

8P3 = 8 × 7 × 6 = 336

Part (b): a committee of 3 is the same committee whichever order the names are written. Order does not matter.

8C3 = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56

Check: each committee can fill the three roles in 3! = 6 ways, and 56 × 6 = 336. The answers agree.

The mistake to watch for

A common slip is to use the permutation because it feels more “complete”.

Question: In how many ways can 4 books be chosen from 9 different books to take on a trip?

Mistaken answer: 9P4 = 9 × 8 × 7 × 6 = 3024

The books are only being chosen, not placed in a sequence, so taking Algebra, Biology, Chemistry, Physics is the same as taking Physics, Chemistry, Biology, Algebra. The 3024 counts each group of 4 books 4! = 24 times.

Correction: 9C4 = 3024 / 24 = 126.

Check yourself

Decide whether order matters first, then calculate.

1. In how many ways can 2 students be chosen from 6 to be monitors with the same duties?

Show answer

Same duties, so order does not matter. 6C2 = (6 × 5) / 2 = 15.

2. Six runners are in a race. In how many ways can gold, silver and bronze be awarded?

Show answer

The three medals are different, so order matters. 6P3 = 6 × 5 × 4 = 120.

3. A coach chooses a team of 5 from 11 players. How many different teams are possible?

Show answer

Only the set of players matters. 11C5 = (11 × 10 × 9 × 8 × 7) / 120 = 55 440 / 120 = 462.

Where this leads next

Once the decision is automatic, move on to counting arrangements with a fixed position, then test yourself on the mixed practice set. The non-calculator working trainer helps with the cancelling.

Some students can do every calculation here but misread which situation a long question describes. A teacher on online one-to-one Additional Mathematics tuition can look at exactly how you read the wording.

Questions people ask

How do I tell quickly whether order matters?

Swap two of the chosen items and ask whether the outcome is now different. If first and second place are different prizes, swapping changes the result, so order matters. If you are only choosing who joins a group, swapping changes nothing, so order does not matter.

What is the link between nPr and nCr?

nPr = nCr × r!. Choosing r items from n gives nCr groups, and each group can be arranged in r! ways. So 8C3 = 56 and 8P3 = 56 × 6 = 336. This also explains why nCr is never larger than nPr.

Do I need a calculator for nPr and nCr?

Both can be done by cancelling, for example 8C3 = (8 × 7 × 6) / 6 = 56. Check the calculator rules for your Additional Mathematics paper on the Cambridge syllabus page, and practise the cancelling method so that you can also verify a calculator answer.

Updated:

Your next step

If you can calculate nPr and nCr but still hesitate over which one a question wants, a one-to-one teacher can listen to how you read the wording and tighten that decision.

Paid one-hour trial at your assigned teacher’s confirmed rate, starting from RM80. Other fees, schedules and ongoing arrangements are confirmed directly with your teacher after the trial class.

Tuition is arranged with a parent or guardian. Send them this page on WhatsApp and they can enquire for you.

Parent or guardian? Enquire here

9,000+ students helped through our service