Permutation vs Combination: When Does Order Matter?
The single most common mistake in probability problems: using combination when permutation is needed, or vice versa. The deciding question is always: does the order of selection matter?
Permutation (nPr): ordered selection — the arrangement ABC differs from BAC. Combination (nCr): unordered selection — {A, B, C} and {B, A, C} are the same.
Formula
nPr = n! / (n−r)! | nCr = n! / (r!(n−r)!)
The Fundamental Difference
The word "arrangement" signals permutation; "selection" or "group" signals combination.
Consider choosing 3 people from a group of 5. If you're picking a President, Vice President, and Treasurer (order matters — different roles), use permutation: P(5,3) = 60. If you're picking a 3-person committee (order doesn't matter — all members equal), use combination: C(5,3) = 10.
The ratio nPr / nCr = r! tells you how many orderings correspond to each combination. For r=3, every combination corresponds to 3! = 6 different permutations.
Decision Guide: Permutation or Combination?
Ask: "If I rearrange the selected items, do I get a different outcome?" YES → permutation. NO → combination.
Use PERMUTATION for: race finishing positions (1st/2nd/3rd), passwords and PIN codes, arrangement of books on a shelf, choosing officers for different roles, seating arrangements.
Use COMBINATION for: lottery ticket selection, choosing a team or committee, quality control sampling, poker hand composition, picking pizza toppings.
| Scenario | Type | Why |
|---|---|---|
| Arrange 5 books in a row | Permutation | Order of books matters — different arrangement = different outcome |
| Pick 3 students for a team | Combination | Team composition matters, not who was picked first |
| Assign gold/silver/bronze medals | Permutation | Each position is distinct |
| Choose 6 lottery numbers from 49 | Combination | Only which numbers; not drawing order |
| Create a 4-digit PIN (no repeat) | Permutation | 1234 ≠ 4321 |
| Select 2 side dishes from 6 options | Combination | Two dishes in any order = same plate |
| Assign roles in a play | Permutation | Actor A as lead ≠ Actor B as lead |
| Choose a sample of 5 items from 100 for testing | Combination | Which 5 tested; not in what order |
The Formulas
Permutation: nPr = n! / (n−r)!. Directly counts ordered sequences. Example: P(10,3) = 10×9×8 = 720.
Combination: nCr = n! / (r! × (n−r)!). Divides permutations by r! to remove order. Example: C(10,3) = 720/6 = 120.
Permutation with repetition (e.g., PINs): nʳ. Example: 4-digit PIN from 10 digits = 10⁴ = 10,000.
Combination with repetition (multiset): C(n+r−1, r). Example: choosing 3 scoops of ice cream from 5 flavors (can repeat) = C(7,3) = 35.
Three Worked Examples
Example 1 (Permutation): How many ways can 4 different colored flags be displayed in a row? All 4 are used: n=4, r=4. P(4,4) = 4! = 24 arrangements.
Example 2 (Combination): A class of 30 students must elect a 4-person student council (all members equal). C(30,4) = 30!/(4!×26!) = 27,405 possible councils.
Example 3 (Mixed): First elect a President from 30 (30 ways), then choose 3 ordinary council members from the remaining 29. Total = 30 × C(29,3) = 30 × 3,654 = 109,620 — this mixes permutation (president) and combination (council members).
Common Mistakes
Mistake 1: Using combination for position problems. "How many ways can 3 runners finish 1st, 2nd, 3rd from 8?" — this is P(8,3) = 336, not C(8,3) = 56.
Mistake 2: Forgetting repeated elements. For anagram problems with repeated letters, divide by the factorial of each repeat. BANANA has 6 letters with B×1, A×3, N×2 → 6! / (1! × 3! × 2!) = 60 distinct anagrams.
Mistake 3: Using nPn when you mean n!. They're equal (nPn = n!/0! = n!×1 = n!), but understanding the formula prevents conceptual errors.