How to Solve Permutations and Combinations? (+FREE Worksheet!)

Permutations and combinations are counting methods used to determine how many ways you can select or arrange items. Permutations are used when order matters; combinations are used when order does not matter. These techniques appear throughout Algebra 1 and are fundamental to probability. This guide explains both formulas step by step, with worked examples and practice problems.

What Are Permutations and Combinations?

Both methods involve choosing items from a larger group, but the key distinction is order:

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• Permutation: An arrangement of items where the order matters. Choosing first, second, and third place in a race is a permutation because 1st, 2nd, 3rd is different from 2nd, 1st, 3rd.
• Combination: A selection of items where the order does NOT matter. Choosing 3 students for a committee is a combination because it does not matter which student was chosen first.

Formulas for Permutations and Combinations

Factorial Notation

The symbol n! (read “n factorial”) means multiply all positive integers from n down to 1.

Quick example: 5! = \(\color{blue}{5 \times 4 \times 3 \times 2 \times 1 = 120}\)    4! = 24    0! = 1 (by definition)

Permutation Formula: P(n, r)

The number of ways to choose and arrange r items from n distinct items (order matters):

\(\color{blue}{P(n, r) = n}\)! ÷ (\(\color{blue}{n – r}\))!

Quick example: \(\color{blue}{P(5, 3) = 5}\)! ÷ (\(\color{blue}{5-3}\))! = 120 ÷ 2 = 58

Combination Formula: C(n, r)

The number of ways to choose r items from n distinct items (order does NOT matter):

\(\color{blue}{C(n, r) = n}\)! ÷ [r! × (\(\color{blue}{n – r}\))!]

Quick example: \(\color{blue}{C(8, 3) = 8}\)! ÷ \(\color{blue}{(3! \times 5!) = 40320}\) ÷ \(\color{blue}{(6 \times 120) = 40320}\) ÷ 720 = 56

Step-by-Step Summary

1. Ask: Does the order of selection matter? Yes → use permutation. No → use combination.
2. Identify n (total items) and r (items to select).
3. Apply the formula: \(\color{blue}{P(n, r) = n}\)! ÷ (\(\color{blue}{n-r}\))! or \(\color{blue}{C(n, r) = n}\)! ÷ [r!(\(\color{blue}{n-r}\))!].
4. Expand and simplify the factorials, then compute the result.

Watch: Permutations and Combinations (Video Lesson)

Shmoop explains permutations and combinations with relatable examples and clear comparisons:

Permutations and Combinations – Worked Examples

Example 1: In how many ways can 3 runners finish 1st, 2nd, and 3rd in a race of 5 runners?

Order matters (1st, 2nd, 3rd are different), so use a permutation.
\(\color{blue}{P(5, 3) = 5}\)! ÷ (\(\color{blue}{5-3}\))! = 120 ÷ 2 = 60 ways

Example 2: A club has 8 members. How many ways can a committee of 3 be chosen?

Order does not matter, so use a combination.
\(\color{blue}{C(8, 3) = 8}\)! ÷ \(\color{blue}{(3! \times 5!) = 40320}\) ÷ \(\color{blue}{(6 \times 120) = 40320}\) ÷ 720 = 56 ways

Example 3: How many 2-letter arrangements can be made from the letters in the word GAMES (no repeats)?

Order matters (AB ≠ BA), so use a permutation. \(\color{blue}{n = 5}\), \(\color{blue}{r = 2}\).
\(\color{blue}{P(6, 2) = 6}\)! ÷ (\(\color{blue}{6-2}\))! = 720 ÷ 24 = 30
Wait — GAMES has 5 letters. \(\color{blue}{P(5, 2) = 5}\)! ÷ 3! = 120 ÷ 6 = 20 arrangements

Example 4: From 10 students, how many ways can a group of 4 be selected for a project?

Order does not matter, so use a combination.
\(\color{blue}{C(10, 4) = 10}\)! ÷ \(\color{blue}{(4! \times 6!) = 3628800}\) ÷ \(\color{blue}{(24 \times 720) = 3628800}\) ÷ 17280 = 210 ways

More Practice: Permutation & Combination Formulas (Video Lesson)

Khan Academy India walks through the formulas with step-by-step derivations:

Exercises: Permutations and Combinations

1. How many ways can you arrange all 4 letters of the word MATH?
2. P(7, 2) = ?
3. C(6, 2) = ?
4. A restaurant offers 5 appetizers and you must choose 2. How many combinations are possible?
5. How many 3-digit PIN codes can be formed from the digits 1–6 with no digit repeated?
6. A team of 3 is chosen from 9 candidates. How many teams are possible?

Answers

1. 4! = 24
2. \(\color{blue}{P(7, 2) = 7}\)! ÷ 5! = 42 → 42
3. \(\color{blue}{C(6, 2) = 6}\)! ÷ \(\color{blue}{(2! \times 4!) = 720}\) ÷ 48 = 15
4. \(\color{blue}{C(5, 2) = 5}\)! ÷ \(\color{blue}{(2! \times 3!) = 120}\) ÷ 12 = 10
5. \(\color{blue}{P(6, 3) = 6}\)! ÷ 3! = 720 ÷ 6 = 120
6. \(\color{blue}{C(9, 3) = 9}\)! ÷ \(\color{blue}{(3! \times 6!) = 362880}\) ÷ 4320 = 84

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Frequently Asked Questions

How do I know whether to use a permutation or a combination?

Ask yourself: does changing the order create a different outcome? If yes (like finishing 1st vs. 2nd in a race, or creating a password), use a permutation. If no (like choosing a team or a committee where positions are not assigned), use a combination.

What is the relationship between P(n, r) and C(n, r)?

\(\color{blue}{C(n, r) = P(n, r)}\) ÷ r! because combinations count each group only once while permutations count every ordering of that same group. Dividing by r! removes the duplicate orderings.

What does 0! equal and why?

By definition, 0! = 1. This is necessary so the formulas work when \(\color{blue}{r = n}\) (choosing all items). It is also consistent with how factorials are derived: n! = \(\color{blue}{n \times (n-1)}\)!, so 1! = \(\color{blue}{1 \times 0}\)! means 0! must equal 1.

Related Topics

• Probability Problems
• Mean, Median, Mode, and Range
• Scatter Plots
• Statistics
• Algebraic Expressions