Permutations vs Combinations: Worked Examples and the Decision Framework
Permutations count arrangements where order matters; combinations count selections where order doesn't. Choosing the wrong formula is the most common error in counting problems. This guide gives the decision framework and 6 worked examples covering passwords, lottery tickets, committee selection, and arrangements with repetition.
What You'll Learn
- ✓Distinguish permutations (order matters) from combinations (order does not matter)
- ✓Apply the formulas P(n,r) = n!/(n−r)! and C(n,r) = n!/[r!(n−r)!]
- ✓Handle problems with repetition allowed vs no repetition
- ✓Recognize hybrid problems requiring both permutation and combination logic
1. Direct Answer: The Decision Framework
The single test: DOES ORDER MATTER? If yes → permutation. If no → combination. Permutation formula: P(n, r) = n! / (n − r)! — counts ordered arrangements of r items from n. Combination formula: C(n, r) = n! / [r! × (n − r)!] — counts unordered selections of r items from n. The key intuition: the combination formula divides the permutation formula by r! to account for the fact that r items can be ordered in r! different ways, but in a combination those r! orderings are all the same selection. Examples: • "Award gold, silver, bronze to 3 of 10 runners" → ORDER MATTERS (different finish positions). Permutation: P(10,3) = 720. • "Choose 3 winners from 10 finalists" → ORDER DOESN'T MATTER. Combination: C(10,3) = 120. • "How many 4-digit codes with no repeated digits?" → ORDER MATTERS. Permutation: P(10,4) = 5040. • "How many 4-card hands from a 52-card deck?" → ORDER DOESN'T MATTER. Combination: C(52,4) = 270,725.
Key Points
- •Permutation: order matters; P(n,r) = n! / (n−r)!
- •Combination: order doesn't matter; C(n,r) = n! / [r!(n−r)!]
- •Single decision test: "If I rearrange the chosen items, is it different?"
- •C(n,r) = P(n,r) / r! — the combination divides out the orderings
- •When in doubt, ask whether the problem cares about position/role/sequence
2. Worked Example 1: Lottery Numbers
A lottery draws 6 numbers from 1-49 without replacement. The order of draw doesn't matter. How many possible tickets? ORDER MATTERS? No (1, 2, 3, 4, 5, 6 is the same ticket as 6, 5, 4, 3, 2, 1). Combination. C(49, 6) = 49! / (6! × 43!) = (49 × 48 × 47 × 46 × 45 × 44) / (6 × 5 × 4 × 3 × 2 × 1) = 13,983,816. The probability of any one ticket being the winning ticket is 1 / 13,983,816 ≈ 0.00000007. About 1 in 14 million. If the lottery had ORDER MATTER (like a Pick-6 sequence), the count would be P(49, 6) = 49! / 43! = 49 × 48 × 47 × 46 × 45 × 44 = 10,068,347,520 — about 10 BILLION possible ordered sequences. The factor of 6! = 720 is exactly the number of orderings of 6 items, which is why combinations are 720x smaller than permutations for r=6.
Key Points
- •Lottery probability is a combination because draw order is irrelevant
- •C(n, r) is much smaller than P(n, r) by a factor of r!
- •A lottery with order-mattering would have 720x more tickets than one without
- •Most real-world lotteries use combinations (order-irrelevant)
3. Worked Example 2: Password Security
How many 8-character passwords using lowercase letters, with repetition allowed? ORDER MATTERS? Yes (ABCDEFGH ≠ HGFEDCBA — both are valid but different passwords). REPETITION ALLOWED? Yes (AAAA1234 is a valid password). When repetition is allowed, the formula is n^r (not the permutation formula). Each of the 8 positions has 26 independent choices. 26^8 = 208,827,064,576. About 209 BILLION possible passwords. Compare to no-repetition (permutation): P(26, 8) = 26! / 18! = 26 × 25 × 24 × ... × 19 = 62,990,928,000. About 63 BILLION. Allowing repetition increases the password space by factor of about 3.3x in this case. With 8 positions and only 26 letters, repetition is essentially required (you couldn't even have an 11-character no-repetition password from just lowercase letters).
Key Points
- •Repetition allowed: use n^r instead of P(n, r)
- •Permutation formula assumes NO repetition by default
- •Order still matters with repetition — passwords ABCD ≠ DCBA
- •Password security calculations use n^r because repetition is typically allowed
4. Worked Example 3: Committee Selection (Combination)
A class of 20 students elects a committee of 5. How many possible committees? ORDER MATTERS? No (the committee is the same set of 5 people regardless of who was selected first). Combination. C(20, 5) = 20! / (5! × 15!) = (20 × 19 × 18 × 17 × 16) / (5 × 4 × 3 × 2 × 1) = 15,504. Now vary the question: "A class of 20 students elects a committee of 5 with specific roles: president, VP, treasurer, secretary, and PR officer. How many?" Now ORDER MATTERS (the same 5 people in different role assignments are different committees). P(20, 5) = 20! / 15! = 20 × 19 × 18 × 17 × 16 = 1,860,480. Note: P(20, 5) = C(20, 5) × 5! = 15,504 × 120 = 1,860,480. The 5! = 120 represents the orderings of 5 people across 5 roles.
Key Points
- •Same group, different question: "committee" (combo) vs "committee with roles" (perm)
- •P(n, r) = C(n, r) × r! — the role assignment multiplies the count by r!
- •Asking "are the roles distinguishable?" is the test for order-matters
- •Real-world voting often uses both: combination for committee election, then permutation for role assignment
5. Worked Example 4: Arrangements with Identical Items
How many distinct arrangements of the letters in MISSISSIPPI? This is a permutation with repeated identical items. Total letters: 11. Distinct letter counts: M=1, I=4, S=4, P=2. Formula: total! / (count1! × count2! × ... × countk!) = 11! / (1! × 4! × 4! × 2!) = 39,916,800 / (1 × 24 × 24 × 2) = 39,916,800 / 1152 = 34,650 distinct arrangements. The naive permutation P(11, 11) = 11! = 39,916,800 would count each arrangement multiple times because rearranging the four I's among themselves doesn't produce a new arrangement. Dividing by 4! × 4! × 2! removes those duplicate counts.
Key Points
- •Repeated identical items require dividing by count factorials
- •Formula: total! / (count of each distinct letter, factorialized)
- •MISSISSIPPI has 11! / (4! × 4! × 2!) = 34,650 arrangements
- •Standard problem in AP Stats and intro probability — memorize the formula
6. Worked Example 5: Hybrid Problem (Both Logics)
A pizza shop offers 12 toppings. How many distinct pizzas can be made with exactly 3 toppings? (Order of toppings doesn't matter — pepperoni-mushroom-onion = onion-pepperoni-mushroom.) ORDER MATTERS? No (toppings are unordered). Combination. C(12, 3) = 12! / (3! × 9!) = (12 × 11 × 10) / (3 × 2 × 1) = 220. Now extend: "How many distinct pizzas using exactly 3 toppings, where the customer can also choose thin or thick crust?" For each topping combination, there are 2 crust options. Total: 220 × 2 = 440. Further extend: "How many distinct 3-topping pizzas using thin or thick crust, in 4 different sizes?" 220 × 2 × 4 = 1,760. Real-world counting problems often combine multiple choices using the multiplication principle: the total count is the product of choices at each independent step.
Key Points
- •Hybrid problems combine combinations, permutations, and the multiplication principle
- •Multiplication principle: independent choices multiply
- •Identify each "layer" of the problem and apply the appropriate formula at each layer
- •Real menus and configuration problems are typically multiplication-principle compositions
7. Worked Example 6: Circular Arrangements
How many distinct ways to seat 8 people around a round table? Intuition: 8! = 40,320 if positions are distinguishable. But around a round table with no fixed "head," rotations of the same arrangement are equivalent. There are 8 rotations of any arrangement, so divide: 8! / 8 = (8 − 1)! = 7! = 5,040 distinct arrangements. General formula: (n − 1)! distinct circular arrangements of n items. If the table also has a "head" or reflective symmetry concerns, the count changes: • Distinguishable head and tail: 8! = 40,320 (linear arrangement) • Round table, no head: 7! = 5,040 (rotations equivalent) • Round table, no head, mirror also considered same: 7!/2 = 2,520 (rotations + reflections) The problem statement determines which version applies. Always read carefully whether positions are distinguishable.
Key Points
- •Circular arrangement of n items: (n − 1)! (rotations equivalent)
- •Linear arrangement of n items: n!
- •Round table with mirror reflections also identified: n!/2
- •Always identify which symmetries the problem treats as equivalent
Key Takeaways
- ★Permutation: order matters; P(n,r) = n! / (n−r)!
- ★Combination: order doesn't matter; C(n,r) = n! / [r!(n−r)!]
- ★C(n,r) = P(n,r) / r! — combinations divide out the r! orderings
- ★Repetition allowed: use n^r instead of P(n, r)
- ★Repeated identical items: total! / (counts factorialized)
- ★Circular arrangement: (n − 1)!
Practice Questions
1. A racing event has 10 runners. How many ways can the gold, silver, and bronze medals be awarded?
2. A poker hand consists of 5 cards from a 52-card deck. How many possible poker hands are there?
3. How many distinct arrangements of the word STATISTICS?
FAQs
Common questions about this topic
Ask: "If I rearrange the chosen items, is it considered a different outcome?" If yes (gold-silver-bronze, codes, sequences) → permutation. If no (committees, hands of cards, sets, lottery tickets) → combination. The phrasing of the question often signals: words like "rank," "arrange," "code," "sequence," "title role" suggest order matters; words like "set," "committee," "team," "selection" suggest order doesn't.
Both involve order mattering, but n^r ALLOWS repetition (each of r positions has all n choices independently) while P(n,r) FORBIDS repetition (each chosen item is removed from the pool). Passwords, license plates, multi-digit codes typically allow repetition. Drawing balls from an urn without replacement, awarding distinct medals, seating people in chairs all forbid repetition. Read the problem to determine which.
Decompose the problem and apply the right formula at each step, then use the multiplication principle. Example: "Choose 5 students from 20, then assign them to 5 distinct roles" = C(20,5) × 5! = P(20,5) = 1,860,480. Alternatively "Choose 5 students, no role assignment" = C(20,5) = 15,504. The multiplication principle is your friend for hybrid problems.
Without replacement: each chosen item is removed from the pool — used in standard P(n,r) and C(n,r) formulas. With replacement: items are drawn and returned, allowing repetition — use n^r for ordered selections. Example: drawing 3 balls from a bag of 10 without replacement: 10 × 9 × 8 = 720 ordered draws. With replacement: 10 × 10 × 10 = 1,000 ordered draws.
Around a round table, an arrangement is the same as any rotation of itself — there's no distinguishable "first" or "last" seat. There are n rotations of every arrangement, so the n! linear arrangements collapse into n!/n = (n-1)! distinct circular arrangements. If reflections (mirror images) are also identified as the same, divide by 2 more: (n-1)!/2.
Yes. Provide the scenario (lottery, password, committee, arrangement, etc.) and StatsIQ identifies whether order matters, whether repetition is allowed, applies the correct formula, and walks through the calculation step by step. Especially useful for distinguishing tricky cases (e.g., hybrid problems with both ordered and unordered components). This content is for educational purposes only and does not constitute statistical advice.