1. **Problem Statement:** (i) Find the number of possible seating arrangements for a family of 10 on a plane with 11 seats. 2. **Formula and Rules:** The number of ways to arrange $n$ distinct people in $k$ distinct seats is given by permutations: $$P(k,n) = \frac{k!}{(k-n)!}$$ 3. **Step (i) Solution:** There are 11 seats and 10 family members. Number of ways = $$P(11,10) = \frac{11!}{(11-10)!} = 11!$$ 4. **Step (ii) Problem:** Given constraints: - 2 grandparents sit in the front row (2 seats) - 4 children each sit at a window seat (4 window seats) - Mr and Mrs Lim sit in the same row on the same side of the aisle - Mr and Mrs Tan sit in another row on the same side of the aisle 5. **Step (ii) Solution:** - Grandparents: 2 seats in front row, arrange 2 grandparents: $$2!$$ - Children: 4 window seats, arrange 4 children: $$4!$$ - Mr and Mrs Lim: same row, same side, 3 seats available, choose 2 seats and arrange couple: $$3 \times 2!$$ - Mr and Mrs Tan: another row, same side, 3 seats available, arrange couple: $$3 \times 2!$$ - Remaining seats for 2 remaining family members: $$2!$$ Total arrangements = $$2! \times 4! \times (3 \times 2!) \times (3 \times 2!) \times 2! = 2 \times 24 \times 6 \times 6 \times 2 = 3456$$ 6. **Step (iii) Problem:** - Mr Lim sits directly behind a child - Mr Tan sits in the front row - Others can sit anywhere 7. **Step (iii) Solution:** - Number of ways to seat Mr Lim behind a child: 4 children in front row window seats, Mr Lim must be behind one of them: 4 choices - Mr Tan sits in front row: 2 front seats, 1 taken by grandparent(s), so 2 choices - Arrange remaining 8 family members in remaining seats: $$8!$$ Total arrangements = $$4 \times 2 \times 8! = 4 \times 2 \times 40320 = 322560$$ --- 8. **Problem 12 (a):** Number of different arrangements of letters in ABRACADABRA (11 letters with repeats) 9. **Formula:** Number of arrangements of letters with repeats: $$\frac{n!}{n_1! n_2! \cdots}$$ 10. **Step (a) Solution:** Letters: A(5), B(2), R(2), C(1), D(1) Number of arrangements = $$\frac{11!}{5! 2! 2! 1! 1!} = 83160$$ 11. **Step (b) Problem:** Find arrangements where: - 2 B's together - 2 R's together - Exactly 4 A's together - C next to D 12. **Step (b) Solution:** Treat pairs/groups as single units: - BB as one unit - RR as one unit - AAAA as one unit - CD as one unit Units: BB, RR, AAAA, CD, plus 1 remaining A Number of units = 5 Number of arrangements = $$5! = 120$$ 13. **Step (c) Problem:** Probability all 5 A's are together 14. **Step (c) Solution:** Treat 5 A's as one unit plus other letters (B,B,R,R,C,D) total 7 units Number of arrangements with 5 A's together = $$\frac{7!}{2! 2!} = 1260$$ Total arrangements = 83160 Probability = $$\frac{1260}{83160} = \frac{1}{66}$$ --- 15. **Additional Practice 1(a):** Number of ways to distribute 5 identical books to 10 people with no one getting more than one 16. **Solution:** Choose 5 people out of 10: $$\binom{10}{5} = 252$$ 17. **Additional Practice 1(b):** Number of ways to distribute 5 different books to 10 people with no restriction 18. **Solution:** Each book can go to any of 10 people: $$10^5 = 100000$$