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Unit 9 · 2.5

Permutations vs combinations

Swap test, pools × , cases +, repeated items — the 4-question guide

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Rule cards

Counting: which tool? Ask these 4 questions in order

Order? → pools? → cases? → repeats? Answer each before you touch the calculator.

  1. 1. Does ORDER matter? Swap two picks — new outcome? Yes → Permutation. No → Combination.
  2. 2. Separate POOLS with quotas (2 men AND 2 women)? → one count per pool, then MULTIPLY.
  3. 3. Several acceptable CASES ('or', 'at least', 'at most')? → count each case, then ADD (or total − the cases you don't want).
  4. 4. Can items REPEAT? Yes (digits, letters of an alphabet) → nʳ. No (people, cards, letters of a WORD) → n × (n − 1) × …
Traps
  • Used a permutation for a plain selection.
    select 3 coins from 6 → P(6,3) = 120
    ✓ C(6,3) = 20 (the key prints 120 — see Key errors)
  • Chose from everyone when the problem set quotas.
    2 men and 2 women → C(12,4)
    ✓ C(men,2) × C(women,2)

Swap test: permutation or combination?

Swap two of the chosen items — if you get a NEW outcome, order matters (permutation); if it's the same group, it's a combination.

P(n, r) and C(n, r) (nPr and nCr, or (n choose r))

  1. Permutation words: titles, 1st/2nd/3rd, codes, passwords, a name, a house number, pictures on a wall, 'assign to different tasks'.
  2. Combination words: committee, team, a selection of books, a collection of DVDs, hand of cards, 'select / choose' with no roles.
  3. Same 4 students: same task → C(12,4) = 495; different tasks → P(12,4) = 11,880.
Traps
  • Treated a committee/selection as ordered.
    plain committee of 3 from 8 → P(8,3) = 336
    ✓ C(8,3) = 336 ÷ 3! = 56
  • Ignored titles or different tasks.
    CEO, director, treasurer from 8 → C(8,3) = 56
    ✓ titles make order matter → P(8,3) = 336
  • Classified the wrong object.
    'a name' → combination because letters are 'selected'
    ✓ swapping letters makes a different name → permutation
TI-84 Evo keys

P(n, r), C(n, r) and why ÷ r!

P(n, r) = n! ÷ (n − r)! = r factors counting down from n; C(n, r) = P(n, r) ÷ r!.

  1. AB and BA are 2 permutations but 1 combination; every group of r shows up r! times → divide by r!.
  2. C(n,1) = n; C(n,r) = C(n,n−r): C(11,7) = C(11,4) = 330.
  3. C(10,3) = (10·9·8) ÷ (3·2·1) = 120.
Traps
  • Multiplied n by r.
    P(4,2) → 4 × 2 = 8
    ✓ P(4,2) = 4·3 = 12 (exam Q23); C(4,2) = 12 ÷ 2 = 6 (exam Q24)
  • Treated 0! as 0 in C(n, n) or C(n, 0).
    C(5,5) = 5!/(5!·0!) → undefined
    ✓ C(n,0) = C(n,n) = 1

Multiply within a case, add across cases

AND = steps that build ONE outcome → ×. OR = separate outcomes with no overlap → +.

  1. List the acceptable cases (2W2M, 3W1M, 4W …).
  2. Inside each case multiply one count per pool.
  3. Add the cases.
  4. Check with total − unwanted cases when 'at least' is involved.
  5. Words CVV + VCV + VVV: 7·5·5 + 5·7·5 + 5·5·5 = 475.
Traps
  • Added the pool counts inside one case.
    3 ice creams and 2 toppings → 120 + 15
    ✓ C(10,3) × C(6,2) = 120 × 15 = 1800
  • Locked in the minimum then chose 'any' for the rest.
    at least 2 women (7W, 5M, choose 4) → C(7,2) × C(10,2) = 945
    ✓ 2W2M 210 + 3W1M 175 + 4W 35 = 420; check C(12,4) − (0W + 1W) = 495 − 75 = 420
  • Left a case out of the sum.
    equal or all one gender → only 2M2W = 90
    ✓ 90 + all men 15 + all women 1 = 106

Arranging with repeated items

Arrange n items with r₁, r₂, … identical copies: n! ÷ (r₁! r₂! …).

  1. AAAB → 4!/3! = 4.
  2. Only the arrangement of ALL the items uses this rule; choosing some of them is a combination problem.
Traps
  • Used n! as if every item were different.
    7 flags → 7! = 5040
    ✓ 3 red, 2 blue, 2 white → 7!/(3!2!2!) = 210
  • Divided by only one of the repeat groups.
    %%%%&&&++ → 9!/4!
    ✓ 9!/(4!3!2!) = 1260
  • Added the repeat factorials in the bottom.
    7!/(3! + 2! + 2!)
    ✓ multiply them: 3!·2!·2! = 24

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