--- title: "Permutations and Combinations" description: "Count arrangements where order matters and selections where it does not. Find the selection count by dividing the arrangement count by the number of orders for each selection." canonical: https://lightmysky.com/learn/mathematics/permutations-and-combinations-mt_1L58kuu-w3 source: https://lightmysky.com/learn/mathematics/permutations-and-combinations-mt_1L58kuu-w3.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Permutations and Combinations Count arrangements where order matters and selections where it does not. Find the selection count by dividing the arrangement count by the number of orders for each selection. Subject: Mathematics · Area: Discrete Mathematics · Ages 18 to 19 Page: https://lightmysky.com/learn/mathematics/permutations-and-combinations-mt_1L58kuu-w3 ## Ready when they can - Choose between a permutation and a combination count from the wording of a problem - Derive the combination formula from the permutation count by dividing out repeats - Count arrangements of a multiset where some objects are identical ## Lesson: Order matters, or it does not Use the product rule for ordered choices and the sum rule for disjoint cases. A permutation distinguishes orders; a combination distinguishes only the chosen group. Test the outcomes, not the word 'choose': choosing people for different posts still distinguishes order. Here P(n,r) and C(n,r) use n distinct objects, with none chosen twice. The integers satisfy 0 <= r <= n. **Example.** Nine students fill two different posts, with nobody holding both. There are 9 choices, then 8, giving 72 assignments. In general, P(n,r) multiplies r decreasing factors: n! / (n-r)!. For positive n, n! means n × (n-1) × ... × 1, and 0! = 1. Each group of r distinct people has r! ordered listings. Divide by that number to count each group once: C(n,r) = P(n,r) / r! = n! / (r! × (n-r)!). For example, choosing 3 people from 8 gives (8 × 7 × 6) / (3 × 2 × 1) = 56. For an unordered pair, the two listings are A then B and B then A. *(drawing: These six lists contain the same members. Distinct posts give six outcomes; equal duties give one group.)* For a row using all n objects, identical copies cause repeated counting. If their type counts are a, b and so on, summing to n, divide n! by a! × b! × ... . Temporarily labelling the copies explains why: each visible row has that many labelled versions. MOM has two Ms and one O, giving 3! / 2! = 3 distinct rows. A required increasing order leaves one listing for each chosen set, so use C(n,r), not P(n,r). Choosing r objects also determines the n-r left out: C(n,r) = C(n,n-r). If one named object must be included, fix it and choose the remaining members from the other n-1 objects. A required member of a pair leaves n-1 choices for its partner. For an arrangement with a fixed first object, remove it before counting the remaining row. Update the type counts used in the divisor. For AABB with A fixed first, the remaining letters are ABB. They give 3! / 2! = 3 rows. The fixed first slot contributes one choice. **Recap.** Distinguish outcomes first. Divide ordered selections by r! for groups, and full arrangements by the factorial of each identical type's count. *Lesson and practice text written for this dataset. Counting resources were consulted for mathematical facts and question-design study.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Counting with the Product and Sum Rules](https://lightmysky.com/learn/mathematics/counting-with-the-product-and-sum-rules-mt_3NZC5F7RKP) ## Opens up - [Binomial Coefficients and Combinatorial Identities](https://lightmysky.com/learn/mathematics/binomial-coefficients-and-combinatorial-identities-mt_lHXFvudTh6)