Exercise 6.4 answers: Permutations and Combinations

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1. Given: ⁿC₈ = ⁿC₂. Find ⁿC₂.
2. When two combination counts from the same n are equal, ⁿCᵣ = ⁿCₛ, either r = s or r + s = n.
3. Here r = 8 and s = 2 are different, so 8 + 2 = n, which gives n = 10.
4. Now find ¹⁰C₂ = 10!/(2!×8!) = (10×9)/2 = 45.

Answer45

Watch this explained “One problem, two routes”, 3:35 into Choosing what to leave out, and the rule that builds each count from two smaller ones

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(i) ²ⁿC₃ : ⁿC₃ = 12 : 1

1. Write ²ⁿC₃ / ⁿC₃ using the combination formula and simplify.
2. ²ⁿC₃ / ⁿC₃ = [2n(2n−1)(2n−2)] / [n(n−1)(n−2)], which simplifies to 4(2n−1)/(n−2).
3. Set this equal to 12: 4(2n−1) = 12(n−2).
4. 8n − 4 = 12n − 24, so 4n = 20, giving n = 5.

Answer5

(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1

1. Use the same simplified ratio: 4(2n−1)/(n−2) = 11.
2. 8n − 4 = 11n − 22, so 3n = 18, giving n = 6.

Answer6

Watch this explained “The sentence as a formula”, 8:27 into Every selection was counted once per arrangement of itself

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1. A chord just needs 2 of the 21 points — the order of the two points does not matter.
2. So the number of chords is ²¹C₂.
3. ²¹C₂ = (21×20)/2 = 210.

Answer210

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1. Choosing the boys and choosing the girls are two separate, independent jobs.
2. Ways to choose 3 boys out of 5: ⁵C₃ = 10.
3. Ways to choose 3 girls out of 4: ⁴C₃ = 4.
4. Since both choices happen together, multiply: 10 × 4 = 40.

Answer40

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. Choosing red, white and blue balls are three separate, independent jobs.
2. Ways to choose 3 red out of 6: ⁶C₃ = 20.
3. Ways to choose 3 white out of 5: ⁵C₃ = 10.
4. Ways to choose 3 blue out of 5: ⁵C₃ = 10.
5. Multiply all three: 20 × 10 × 10 = 2000.

Answer2000

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 ace out of the 4 aces: ⁴C₁ = 4.
2. The other 4 cards must come from the remaining 48 non-ace 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. Choose exactly 4 bowlers out of the 5 who can bowl: ⁵C₄ = 5.
2. The remaining 7 team members come from the 12 non-bowlers: ¹²C₇ = 792.
3. Multiply the two independent choices: 5 × 792 = 3960.

Answer3960

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 2 black balls out of 5: ⁵C₂ = 10.
2. Choose 3 red balls out of 6: ⁶C₃ = 20.
3. Multiply the two independent choices: 10 × 20 = 200.

Answer200

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. The 2 compulsory courses are already fixed as part of every programme.
2. So the student really only chooses the remaining 5 − 2 = 3 courses.
3. These 3 courses come from the 9 − 2 = 7 courses that are left: ⁷C₃ = 35.

Answer35

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