How to Calculate Permutations and Combinations

Permutations and combinations are the mathematical tools for counting arrangements and selections. Whether you are calculating lottery odds, determining how many teams can be formed from a group, or figuring out the number of possible passwords, these two concepts provide the answers.

This guide explains both concepts from the ground up, covers the factorial operations that power the formulas, and walks through real-world counting problems step by step.

What Are Permutations and Combinations?

Both permutations and combinations deal with selecting items from a larger set, but they differ in one critical way: whether order matters.

Think of it this way: a permutation cares about the arrangement of items, while a combination only cares about the selection. Picking a president, vice president, and treasurer from a club of 10 people is a permutation problem (each role is different). Picking 3 people from the same club to form a committee with no specific roles is a combination problem.

Understanding Factorials

Both formulas rely on the factorial operation, written as n! (read "n factorial"). The factorial of a positive integer n is the product of all positive integers from 1 to n.

Factorials grow astronomically fast. While 10! is about 3.6 million, 20! exceeds 2.4 quintillion. This rapid growth reflects the explosion of possible arrangements as the number of items increases and explains why large counting problems produce such enormous results.

The Permutation Formula

The number of permutations of r items selected from a set of n items is given by:

The formula works by recognizing that for the first position you have n choices, for the second you have n - 1 remaining choices, for the third n - 2, and so on for r positions. The product n × (n - 1) × (n - 2) × ... × (n - r + 1) equals n! divided by (n - r)!.

The Combination Formula

The number of combinations of r items selected from n items is:

The combination formula is the permutation formula divided by r!, which eliminates the duplicate orderings within each group. Since r items can be arranged in r! different orders, dividing by r! counts each group only once.

When to Use Permutations vs Combinations

The single question that determines which formula to use is: does the order of selection matter? Here is a decision guide:

A helpful test: imagine swapping two selected items. If the swap creates a different outcome (swapping who gets gold vs. silver), use permutations. If the swap changes nothing (swapping which two people are on the committee), use combinations.

Practical Examples

Counting problems appear everywhere from games of chance to logistics planning. These examples demonstrate how permutations and combinations solve real questions.

Counting Reference Table

This table provides quick reference values for common permutation and combination calculations. Notice how permutations always exceed or equal combinations for the same n and r values.

Tips and Complete Guide

Counting problems become manageable when you follow a systematic approach. These strategies will help you identify the right formula and avoid errors.

• Always ask "does order matter?" first. This single question determines whether you use permutations or combinations. If rearranging the selected items creates a different outcome, use permutations. If not, use combinations.
• Simplify before multiplying. In the formula C(49, 6), cancel common factors between the numerator and denominator before multiplying everything out. Compute (49 × 48 × 47 × 46 × 45 × 44) / 720 rather than calculating full factorials.
• Use symmetry to check answers. C(n, r) always equals C(n, n - r). So C(10, 3) = C(10, 7) = 120. If your answer does not satisfy this property, recheck your work.
• Draw tree diagrams for small problems. When n and r are small (3 or 4 items), sketching a tree diagram shows every possibility and helps you verify your formula-based answer.
• Break complex problems into stages. If you need to choose 2 guards and 3 forwards from a pool of 6 guards and 8 forwards, calculate each selection independently: C(6, 2) × C(8, 3) = 15 × 56 = 840. This uses the fundamental counting principle to combine independent choices.
• Watch for restrictions. Problems that say "at least one" or "at most three" often require calculating the complementary case and subtracting. The number of committees with at least one woman equals the total committees minus the all-male committees.

Common Mistakes to Avoid

• Confusing permutations and combinations. Using P(n, r) when order is irrelevant inflates your answer by a factor of r!. A 5-card poker hand has C(52, 5) = 2,598,960 possible hands, not P(52, 5) = 311,875,200. Always determine whether ordering matters before selecting a formula.
• Forgetting that 0! = 1. When r = n (selecting all items), the formula includes (n - n)! = 0! in the denominator. If you treat 0! as 0 instead of 1, you will get a division-by-zero error rather than the correct result.
• Misidentifying repetition. Standard permutation and combination formulas assume each item is selected at most once. If items can repeat (like digits in a PIN code), you need the "with repetition" formulas: n^r for permutations or C(n + r - 1, r) for combinations.
• Computing full factorials unnecessarily. Calculating 49! as a complete number before dividing is impractical and risks overflow errors. Cancel the common factorial terms first, then multiply the remaining factors.
• Ignoring problem constraints. Read problems carefully for restrictions like "must include at least 2 seniors" or "no two people from the same department." These constraints require adjusting the calculation, often by splitting into cases or using complementary counting.

Last updated: February 23, 2026