Permutations — Solutions

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1. Q1 — Calculate Permutations

   Fluency
   • (a) P(6, 2) = 6 × 5 = 30.
   • (b) P(9, 3) = 9 × 8 × 7 = 504.
   • (c) P(5, 5) = 5! = 120.

2. Q2 — Seating Arrangements

   Fluency
   • 8 people in 8 positions: P(8, 8) = 8! = 40 320 ways.

3. Q3 — 3-Letter Arrangements

   Fluency
   • Order matters, no repetition. P(7, 3) = 7 × 6 × 5 = 210 arrangements.

4. Q4 — Selecting Office-Bearers

   Fluency
   • Three distinct roles from 12 people. P(12, 3) = 12 × 11 × 10 = 1 320 ways.

5. Q5 — Arranging Letters

   Understanding
   • (a) MATHS — 5 distinct letters: 5! = 120.
   • (b) MISSISSIPPI — 11 letters: M(1), I(4), S(4), P(2). Arrangements = 11!/(4! × 4! × 2!) = 34 650.

6. Q6 — Row Seating with Restriction

   Understanding
   • (a) No restrictions: 6! = 720 ways.
   • (b) Ann and Bob adjacent: treat as one unit → 5 units. 5! × 2 = 120 × 2 = 240 ways.

7. Q7 — 5-Digit Numbers from {1, 2, 3, 4, 5}

   Understanding
   • (a) Total: 5! = 120.
   • (b) Even (last digit 2 or 4): 2 × 4! = 48.
   • (c) 3 fixed in position 3: 4! = 24.

8. Q8 — Arrangements of PARALLEL

   Understanding
   • PARALLEL: 8 letters, A(2), L(3). Arrangements = 8!/(2! × 3!) = 40 320/12 = 3 360.

9. Q9 — Bookshelf Arrangements

   Problem Solving
   • (a) No restrictions: 9! = 362 880 ways.
   • (b) Groups together: 3! subject blocks × 4! × 3! × 2! internal = 6 × 24 × 6 × 2 = 1 728 ways.

10. Q10 — 4-Digit Numbers with Leading Digit Restriction

    Problem Solving
    • Leading digit from {1,2,3,4,5}: 5 choices. Remaining 3 positions from remaining 5 digits: P(5,3) = 60. Total = 5 × 60 = 300 four-digit numbers.