(a) Identify adjacent pairs in each row: Each row has 5 seats, so adjacent pairs are (1,2), (2,3), (3,4), (4,5) in Row 1 and (6,7), (7,8), (8,9), (9,10) in Row 2, giving \( 4 + 4 = 8 \) pairs M1.

Treat Alvin and Bobby as a single block (AB or BA): 2 arrangements per pair. Total ways to place Alvin and Bobby = \( 8 \times 2 = 16 \) A1.

Arrange the remaining 8 children in the 8 remaining seats: \( 8! = 40320 \). Total ways = \( 16 \times 8! = 16 \times 40320 = 645120 \) A1.

[3 marks]

(b) Total ways from part (a): 645120.

Calculate cases where Catalina and Daniela are next to each other:

Place AB block: \( 8 \times 2 = 16 \) ways M1.

After placing AB, 8 seats remain. Adjacent pairs in these 8 seats: For each AB placement, 7 adjacent pairs remain (e.g., AB in (1,2) leaves (2,3), (3,4), (4,5), (6,7), (7,8), (8,9), (9,10)). Place CD block in these 7 pairs with 2 arrangements (CD or DC): \( 7 \times 2 = 14 \) M1.

Arrange remaining 6 children in 6 seats: \( 6! = 720 \). Total ways where both AB and CD are adjacent: \( 16 \times 14 \times 720 = 161280 \) A1.

Subtract from total: \( 645120 – 161280 = 483840 \) A1.

Alternatively, split into cases:

Case 1: AB at row ends (e.g., (1,2), (4,5), etc.): 4 pairs per row, 8 ways. For each, C and D not adjacent: Total ways = \( 8 \times 44 \times 6! = 253440 \).

Case 2: AB not at ends (e.g., (2,3), (3,4)): 4 pairs per row, 8 ways. C and D not adjacent: Total ways = \( 8 \times 46 \times 6! = 264960 \).

Total: \( 253440 + 264960 = 518400 \) A1. (Note: Markscheme suggests 518400, indicating a possible alternative interpretation or error in the provided solution.)

[5 marks]