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

Order matters in a permutation but not in a combination. Use this distinction when counting arrangements or selections.

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 = 60

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

What is the difference between permutations and combinations?

In a permutation the order matters; in a combination it does not. Choosing a president and a vice president from 5 students is a permutation, because Ana-then-Ben differs from Ben-then-Ana, so there are 5 × 4 = 20 outcomes. Choosing a 2-person committee from the same 5 students is a combination, and there are only 20 ÷ 2 = 10 possible groups.

What is the formula for permutations and combinations?

Permutations use nPr = n! ÷ (n − r)!, and combinations use nCr = n! ÷ (r!(n − r)!). Here n is the number of items and r is how many you pick. For 6 items taken 2 at a time, 6P2 = 720 ÷ 24 = 30, while 6C2 = 30 ÷ 2 = 15, because each pair would otherwise be counted twice.

How do you identify permutations and combinations?

Ask whether swapping two chosen items gives a new result. If it does, use a permutation; if not, use a combination. Words like arrange, rank, order, line up, or code usually signal permutations, while choose, select, group, team, or committee usually signal combinations. A 4-digit PIN is a permutation problem; picking 3 toppings for a pizza is a combination problem.

How do you do permutations and combinations on a TI-84?

Type n, press MATH, arrow right to PRB, and choose nPr (option 2) or nCr (option 3), then type r and press ENTER. For example, 7, nCr, 3 gives 35, the number of ways to choose 3 books from 7. Using 7, nPr, 3 gives 210, because order now counts.

What grade level is permutations and combinations?

Most students meet formal permutations and combinations in high school, usually in Algebra 2, precalculus, or a statistics course. The groundwork comes earlier: middle school students count outcomes with lists, tree diagrams, and the fundamental counting principle, such as 3 shirts × 4 pants = 12 outfits.

How do you use permutations and combinations in probability?

Count the favorable outcomes and the total outcomes with the same method, then divide. Suppose a bag holds 3 red and 2 blue marbles and you draw 2. There are 5C2 = 10 possible pairs and 3C2 = 3 red pairs, so P(both red) = 3/10. Order does not matter here, so combinations keep the counting simple.

Related Topics

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