Pascal’s Triangle and Combinations — Solutions

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1. Evaluate each combination. Fluency

   1. (a) ⁵C₂ = 5!/(2!3!) = (5×4)/(2×1) = 10
   2. (b) ⁷C₀ = 1 (choosing 0 items — only one way to choose nothing)
   3. (c) ⁶C₆ = 1 (choosing all 6 — only one way)
   4. (d) ⁸C₃ = 8!/(3!5!) = (8×7×6)/(3×2×1) = 336/6 = 56
   5. (e) ¹⁰C₄ = 10!/(4!6!) = (10×9×8×7)/(4×3×2×1) = 5040/24 = 210
   6. (f) ⁹C₇ By symmetry: ⁹C₇ = ⁹C₂ = (9×8)/2 = 36

2. Pascal’s triangle. Fluency

   1. (a) Row 6. 1, 6, 15, 20, 15, 6, 1
   2. (b) Row 7 using Pascal’s rule. Each entry = sum of two entries above it in row 6:
      1, 7, 21, 35, 35, 21, 7, 1
   3. (c) Row sums. Row 5 sum: 1+5+10+10+5+1 = 32 = 2&sup5;.
      Row 6 sum: 1+6+15+20+15+6+1 = 64 = 2&sup6;.
      Pattern: the sum of row n is always 2ⁿ.

3. Symmetry property. Fluency

   1. (a) ¹⁰C₈ = ¹⁰C₂ = (10×9)/2 = 45
   2. (b) ¹²C₁₀ = ¹²C₂ = (12×11)/2 = 66
   3. (c) ¹⁵C₁₃ = ¹⁵C₂ = (15×14)/2 = 105
   4. (d) ²⁰C₁₈ = ²⁰C₂ = (20×19)/2 = 190

4. Factorial expressions. Fluency

   1. (a) 5!/3! = 5×4×3!/3! = 5×4 = 20
   2. (b) 8!/(6!×2!) = (8×7)/(2×1) = 56/2 = 28 = ⁸C₂
   3. (c) (n+1)!/n! = (n+1) × n! / n! = n+1
   4. (d) n!/(n−2)! = n × (n−1) × (n−2)! / (n−2)! = n(n−1)

5. Pascal’s rule applications. Fluency

   1. (a) ⁹C₄ ⁹C₄ = ⁸C₃ + ⁸C₄ = 56 + 70 = 126
   2. (b) ⁹C₅ ⁹C₅ = ⁹C₄ = 126 (by symmetry: ⁹C₅ = ⁹C₄)
   3. (c) ¹⁰C₅ ¹⁰C₅ = ⁹C₄ + ⁹C₅ = 126 + 126 = 252

6. Solve for the unknown. Understanding

   1. (a) ⁿC₂ = 15. n(n−1)/2 = 15 → n(n−1) = 30 → n²−n−30 = 0 → (n−6)(n+5) = 0.
      n = 6 (taking positive value).
   2. (b) ⁿC₁ = 7. ⁿC₁ = n = 7.
   3. (c) ⁶C_r = 15. From row 6: 1, 6, 15, 20, 15, 6, 1. The value 15 appears at positions r = 2 and r = 4 (by symmetry).
   4. (d) ⁿC₃ = 10. n(n−1)(n−2)/6 = 10 → n(n−1)(n−2) = 60.
      Try n = 5: 5×4×3 = 60 ✓. So n = 5.

7. Counting selections. Understanding

   1. (a) Team of 4 from 9. ⁹C₄ = 126 teams.
   2. (b) Committee of 3 with exactly 2 women. Choose 2 women from 4: ⁴C₂ = 6.
      Choose 1 man from 5: ⁵C₁ = 5.
      Total: 6 × 5 = 30 committees.
   3. (c) 5-card hands with exactly 3 aces. Choose 3 aces from 4: ⁴C₃ = 4.
      Choose 2 non-aces from 48: ⁴⁸C₂ = (48×47)/2 = 1128.
      Total: 4 × 1128 = 4512 hands.

8. Properties of Pascal’s triangle. Understanding

   1. (a) Row sums = 2ⁿ. n=3: 1+3+3+1 = 8 = 2³ ✓
      n=4: 1+4+6+4+1 = 16 = 2&sup4; ✓
      n=5: 1+5+10+10+5+1 = 32 = 2&sup5; ✓
   2. (b) Alternating sum. n=3: 1−3+3−1 = 0.
      n=4: 1−4+6−4+1 = 0.
      n=5: 1−5+10−10+5−1 = 0.
      Pattern: the alternating sum is always 0 for n ≥ 1.

9. Proving combination identities. Problem Solving

   1. (a) Prove ⁿC_r = ⁿC_n−r. ⁿC_n−r = n!/((n−r)!(n−(n−r))!) = n!/((n−r)!r!) = n!/(r!(n−r)!) = ⁿC_r ✓
   2. (b) Prove Pascal’s rule. ⁿ⁻¹C_r−1 + ⁿ⁻¹C_r
      = (n−1)!/((r−1)!(n−r)!) + (n−1)!/(r!(n−1−r)!)
      Common denominator r!(n−r)!:
      = (n−1)! × r / (r!(n−r)!) + (n−1)! × (n−r) / (r!(n−r)!)
      = (n−1)![r + (n−r)] / (r!(n−r)!)
      = (n−1)! × n / (r!(n−r)!)
      = n! / (r!(n−r)!) = ⁿC_r ✓

10. Advanced counting. Problem Solving

    1. (a) Dividing 12 into two unlabelled groups of 6. Total ways to choose one group of 6 from 12: ¹²C₆ = 924.
       But each split is counted twice (group A vs group B is the same as group B vs group A).
       Answer: 924/2 = 462 ways.
    2. (b) President, VP, and 3-person committee from 10. Choose president: 10 ways. Choose VP: 9 ways. (These two cannot be on committee.)
       Choose 3-person committee from remaining 8: ⁸C₃ = 56.
       Total: 10 × 9 × 56 = 5040 ways.
    3. (c) Diagonals of a convex polygon. ⁿC₂ counts all pairs of vertices (including edges). Subtracting n edges gives only diagonals: D = ⁿC₂ − n.
       Decagon (n=10): D = ¹⁰C₂ − 10 = 45 − 10 = 35 diagonals.