5 min read
2026-02-05
The main difficulty in combinatorics is determining which formula to apply. These tips will help you figure it out.
**Does order matter?** If yes — it's arrangements or permutations. If no — combinations.
**Are all elements used?** If yes — permutations. If selecting a subset — combinations or arrangements.
How many ways to arrange 5 books on a shelf? → Permutations: P(5) = 120
How many ways to choose 3 books from 10? → Combinations: C(10,3) = 120
How many three-digit codes from digits 1-9? → Arrangements: A(9,3) = 504
If a problem consists of several independent stages, multiply the number of options at each stage.
The result cannot be greater than the total number of all possible permutations. If C(n,k) > P(n), there's an error somewhere.
Arrangements with repetition (n^k) and without repetition (A(n,k)) use different formulas. Determine whether repeated selection is allowed.
C(n,k) = C(n, n-k). Choosing 2 from 10 is the same as discarding 8 from 10. This simplifies calculations for large k.
See also: Prime Number Checker, Statistics, Equation Solver