Permutations vs. Combinations: Does Order Matter?

Counting problems ask “how many ways?” How many passwords, how many teams, how many orders of finishing a race. The answer comes down to one question you ask every time: does the order matter? If it does, it is a permutation. If it doesn’t, it is a combination.

The fundamental counting principle

If one choice can be made in m ways and a second in n ways, both can be made in m × n ways. This extends to any number of choices.

Example A restaurant offers 4 starters, 6 mains and 3 desserts. How many three-course meals are possible?

4 × 6 × 3 = 72.

Example How many 4-digit PINs are possible if digits can repeat?

Each of the 4 positions has 10 options: 10 × 10 × 10 × 10 = 10,000.

Example How many 4-digit PINs have no repeated digit?

10 options, then 9, then 8, then 7: 10 × 9 × 8 × 7 = 5,040.

Draw a blank for each position, write the number of options in each, and multiply. That method alone solves most counting problems on the SAT and ACT.

Factorials

n! (“n factorial”) is n × (n − 1) × … × 2 × 1. It counts the ways to arrange n different things in a row.

  • 3! = 6
  • 5! = 120
  • 10! = 3,628,800
  • 0! = 1 by definition (there is exactly one way to arrange nothing).

Example In how many orders can 6 runners finish a race (no ties)?

6! = 720.

Permutations: order matters

A permutation is an arrangement where order matters. The number of ways to pick and arrange r things from n is:

nPr = n!/(n − r)!

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

  1. Order matters, because being president differs from being treasurer.
  2. 12P3 = 12!/9! = 12 × 11 × 10 = 1,320.

The blank-and-multiply method gives the same thing: 12 choices for president, then 11, then 10.

Combinations: order doesn’t matter

A combination is a selection where order doesn’t matter. Choosing Ana, Ben and Cal for a team is the same team as Cal, Ana and Ben. The formula is:

nCr = n!/(r!(n − r)!)

It is also written as (n choose r). It equals the permutation count divided by r!, because each group of r can be arranged r! ways and we only want to count it once.

Example A club of 12 people chooses a 3-person committee. How many committees are possible?

  1. Order doesn’t matter: there are no roles.
  2. 12C3 = 1,320/3! = 1,320/6 = 220.

Quick test: swap two people

If swapping two chosen people gives a different outcome, order matters (permutation). If it gives the same outcome, it doesn’t (combination).

Side by side

QuestionOrder matters?Count
Top 3 finishers (1st, 2nd, 3rd) of 8 runnersyes8P3 = 336
3 runners of 8 who qualify for the finalno8C3 = 56
5-card hands from a 52-card deckno52C5 = 2,598,960
4-letter codes from 26 letters, no repeatsyes26P4 = 358,800

Arrangements with repeated items

When some items are identical, divide by the factorial of each repeat count, because swapping identical items doesn’t create a new arrangement.

Example How many distinct arrangements of the letters in BANANA are there?

  1. 6 letters: B ×1, A ×3, N ×2.
  2. 6!/(3! × 2!) = 720/12 = 60.

Combinations with conditions

Break the selection into parts and multiply.

Example A committee of 4 is chosen from 7 juniors and 5 seniors. How many committees have exactly 2 seniors?

  1. Choose 2 of 5 seniors: 5C2 = 10.
  2. Choose 2 of 7 juniors: 7C2 = 21.
  3. Multiply: 10 × 21 = 210.

For “at least one” questions, it is usually easier to count the total and subtract the cases with none.

Example How many 4-person committees from the same group have at least one senior?

  1. All committees: 12C4 = 495.
  2. Committees with no seniors (all juniors): 7C4 = 35.
  3. At least one senior: 495 − 35 = 460.

Probability with combinations

Probability is favorable outcomes over total outcomes, and combinations count both. The probability basics guide covers the rules this builds on.

Example A bag has 5 red and 4 blue marbles. Three are drawn at random. What is the probability all three are red?

  1. Ways to choose 3 red: 5C3 = 10.
  2. Ways to choose any 3: 9C3 = 84.
  3. Probability: 10/84 = 5/42 ≈ 0.119.

On a calculator, nPr and nCr are under MATH → PRB on a TI-84. In Desmos, type nPr(12,3) or nCr(12,3).

Your turn

  1. How many license plates have 3 letters followed by 3 digits, with repeats allowed?

    Show answer

    17,576,000. 26³ × 10³ = 17,576 × 1,000.

  2. In how many ways can 5 books be arranged on a shelf?

    Show answer

    120. 5! = 120.

  3. How many ways can a coach choose 5 starters from a team of 11?

    Show answer

    462. Order doesn’t matter: 11C5 = 462.

  4. How many ways can gold, silver and bronze be awarded among 10 finalists?

    Show answer

    720. Order matters: 10 × 9 × 8.

  5. How many distinct arrangements of the letters in LETTER are there?

    Show answer

    180. 6 letters with E ×2 and T ×2: 720/(2 × 2).

  6. Two cards are drawn from a standard deck without replacement. What is the probability both are hearts?

    Show answer

    1/17. 13C2/52C2 = 78/1,326 = 1/17.

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Drafted with help from AI tools. Every example is worked through step by step so you can check it.