Combinations vs. Permutations
A combination represents the number of ways to choose r items from n total items where the order doesn't matter. If order becomes relevant—such as arranging people in specific positions—you're working with permutations instead.
The key distinction: selecting 3 people from 10 for a committee gives one combination, but arranging those same 3 people into president, vice-president, and treasurer roles produces multiple permutations. Combinations ignore sequence; permutations depend on it.
In practical terms:
- Combinations: "Which 5 lottery numbers did I pick?" (order irrelevant)
- Permutations: "What was the winning sequence?" (order critical)
Combination Formula
The standard combination formula calculates selections without repetition. When items can be chosen multiple times, a modified formula accounts for the increased possibilities.
C(n,r) = n! ÷ (r! × (n − r)!)
C(n,r) with repetition = (r + n − 1)! ÷ (r! × (n − 1)!)
n— Total number of distinct objects availabler— Number of objects you want to select!— Factorial: the product of all positive integers up to that number (e.g., 5! = 5 × 4 × 3 × 2 × 1 = 120)
Common Pitfalls and Considerations
Understanding when and how to apply combination formulas prevents calculation mistakes.
- Confusing combinations with permutations — Many people mistakenly use permutation formulas when order is irrelevant. For selecting 5 committee members from 20 candidates, use combinations—not permutations. The permutation result will be 24 times larger because it treats different orderings as distinct outcomes.
- Forgetting to account for repetition rules — Standard combinations assume each item can only be selected once. If you're choosing with replacement—like picking coloured balls returned to the urn each time—use the repetition formula instead. Ignoring this distinction produces dramatically different answers.
- Handling large factorials — Factorials grow extremely fast: 20! exceeds 2 trillion. Most calculators express results in scientific notation (mantissa and exponent) to avoid display overflow. A result like 1.23 × 10^15 means 1,230,000,000,000,000 exact combinations.
- Verifying when n equals r — When selecting all items (r = n), there is exactly one combination. This often surprises users, but it makes sense: there's only one way to choose everything. The formula correctly produces C(n,n) = 1 in all cases.
Worked Example: 5-Letter Combinations
Suppose you need to find how many distinct 5-letter combinations exist using the 26-letter English alphabet.
Step 1: Identify your variables. n = 26 (total letters), r = 5 (letters per combination).
Step 2: Apply the formula without repetition:
C(26,5) = 26! ÷ (5! × 21!) = 65,780
Step 3: Interpret the result. There are 65,780 unique 5-letter groupings. If repetition were allowed (selecting the same letter multiple times), the formula yields 142,506 combinations—roughly double.
Most practical problems involve selections without repetition: lottery draws, committee formation, or test question selection. Repetition models apply mainly to sampling-with-replacement scenarios or password generation.