Miscellaneous Exercise answers: Permutations and Combinations

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1. DAUGHTER has 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R), all different letters.
2. Choose 2 vowels out of 3: ³C₂ = 3.
3. Choose 3 consonants out of 5: ⁵C₃ = 10.
4. Now arrange the 5 chosen letters in a row: 5! = 120.
5. Multiply: 3 × 10 × 120 = 3600.

Answer3600

Watch this explained “Read it right to left”, 5:03 into Every selection was counted once per arrangement of itself

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1. EQUATION has 8 different letters: 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N).
2. Tie all 5 vowels into one block and all 3 consonants into another block.
3. Arrange the 2 blocks: 2! = 2 ways.
4. Arrange the 5 vowels inside their block: 5! = 120 ways.
5. Arrange the 3 consonants inside their block: 3! = 6 ways.
6. Multiply: 2 × 120 × 6 = 1440.

Answer1440

Watch this explained “Keeping a group together”, 13:13 into The closed formula, and how allowing repeats changes the count entirely

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(i) exactly 3 girls

1. Exactly 3 girls means the other 7 − 3 = 4 members are boys.
2. Choose 3 girls out of 4: ⁴C₃ = 4.
3. Choose 4 boys out of 9: ⁹C₄ = 126.
4. Multiply: 4 × 126 = 504.

Answer504

(ii) atleast 3 girls

1. There are only 4 girls in all, so "at least 3 girls" means exactly 3 girls or exactly 4 girls. These two cases never overlap, so we add them.
2. Exactly 3 girls (from part (i)): 504.
3. Exactly 4 girls: choose all 4 girls in ⁴C₄ = 1 way, and the other 3 members from the 9 boys in ⁹C₃ = 84 ways, giving 1 × 84 = 84.
4. Add the two cases: 504 + 84 = 588.

Answer588

(iii) atmost 3 girls

1. "At most 3 girls" means 0, 1, 2 or 3 girls. It is quicker to count the committees we do not want and subtract.
2. All committees of 7 from 9 + 4 = 13 people: ¹³C₇ = 1716.
3. The only committees not allowed are those with exactly 4 girls: 84 (from part (ii)).
4. So the committees with at most 3 girls number 1716 − 84 = 1632.

Answer1632

Watch this explained “When you may multiply”, 8:47 into Choosing what to leave out, and the rule that builds each count from two smaller ones

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1. EXAMINATION has 11 letters: A and I and N each repeat twice, the rest (E, X, M, T, O) appear once.
2. In dictionary order, the letters sorted alphabetically are A, A, E, I, I, M, N, N, O, T, X — the only letter before E is A.
3. So every word that comes before the first word starting with E is a word that starts with A.
4. Fix one A at the front. The remaining 10 letters are A, E, X, M, I, I, N, N, T, O — with I and N still repeated twice each.
5. Arrange these 10 letters: 10!/(2!×2!) = 3628800/4 = 907200.

Answer907200

Watch this explained “Restrictions on top”, 8:50 into Dividing out the swaps you cannot see when some objects are identical

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1. A number is divisible by 10 only if its last digit is 0, so 0 must go in the last place.
2. The remaining 5 digits (1, 3, 5, 7, 9) fill the other 5 places, none of them zero, so there's no leading-zero problem.
3. Arrange these 5 digits in the 5 remaining places: 5! = 120.

Answer120

Watch this explained “When you use everything”, 9:26 into Filling a row of places one at a time from a supply of distinct objects

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1. Choose 2 different vowels out of 5: ⁵C₂ = 10.
2. Choose 2 different consonants out of 21: ²¹C₂ = 210.
3. Arrange the 4 chosen letters in a row: 4! = 24.
4. Multiply: 10 × 210 × 24 = 50400.

Answer50400

Watch this explained “Read it right to left”, 5:03 into Every selection was counted once per arrangement of itself

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1. 8 questions must be split between Part I (5 questions) and Part II (7 questions), with at least 3 from each part.
2. The possible splits are 3 from Part I with 5 from Part II, 4 with 4, or 5 with 3 — these three cases never overlap.
3. 3 from Part I, 5 from Part II: ⁵C₃ × ⁷C₅ = 10 × 21 = 210.
4. 4 from Part I, 4 from Part II: ⁵C₄ × ⁷C₄ = 5 × 35 = 175.
5. 5 from Part I, 3 from Part II: ⁵C₅ × ⁷C₃ = 1 × 35 = 35.
6. Add the three cases: 210 + 175 + 35 = 420.

Answer420

Watch this explained “When you may multiply”, 8:47 into Choosing what to leave out, and the rule that builds each count from two smaller ones

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1. Choose exactly 1 king out of the 4 kings: ⁴C₁ = 4.
2. The other 4 cards must come from the remaining 48 non-king cards: ⁴⁸C₄ = 194580.
3. Multiply the two independent choices: 4 × 194580 = 778320.

Answer778320

Watch this explained “When you may multiply”, 8:47 into Choosing what to leave out, and the rule that builds each count from two smaller ones

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1. There are 5 + 4 = 9 seats in the row, numbered 1 to 9.
2. The even-numbered seats are 2, 4, 6, 8 — exactly 4 of them, matching the 4 women.
3. Arrange the 4 women in these 4 even seats: 4! = 24 ways.
4. Arrange the 5 men in the 5 remaining (odd-numbered) seats: 5! = 120 ways.
5. Multiply: 24 × 120 = 2880.

Answer2880

Watch this explained “When you use everything”, 9:26 into Filling a row of places one at a time from a supply of distinct objects

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1. The 3 special students act as a single unit: either all 3 join, or none of them do — these two cases never overlap.
2. If all 3 join: the other 7 spots come from the remaining 25 − 3 = 22 students: ²²C₇ = 170544.
3. If none of the 3 join: all 10 spots come from the remaining 22 students: ²²C₁₀ = 646646.
4. Add the two cases: 170544 + 646646 = 817190.

Answer817190

Watch this explained “In, or out”, 6:32 into Choosing what to leave out, and the rule that builds each count from two smaller ones

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1. ASSASSINATION has 13 letters: S appears 4 times, A 3 times, I 2 times, N 2 times, and T and O once each.
2. Tie all 4 S's into one block. Now there are 9 remaining letters (A×3, I×2, N×2, T, O) plus 1 block — 10 items to arrange.
3. Since the S's are identical, there's no separate arrangement needed inside the block.
4. Arrange the 10 items, dividing by the repeats among A, I and N: 10!/(3!×2!×2!) = 3628800/24 = 151200.

Answer151200

Watch this explained “Restrictions on top”, 8:50 into Dividing out the swaps you cannot see when some objects are identical