a(i). [1 mark]

With no restrictions, arrange the 9 team members (6 boys + 3 girls) in a line: \( 9! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 362880 \) ways (A1).

a(ii). [2 marks]

Treat the three girls as a single entity (block) (M1).
Arrange 7 entities (6 boys + 1 girl-block): \( 7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040 \).
Arrange the 3 girls within the block: \( 3! = 6 \) ways.
Total ways: \( 7! \cdot 3! = 5040 \cdot 6 = 30240 \) (A1).

b. [3 marks]

Find selections of 5 members with at least 2 girls (M1):
Case 1: Exactly 2 girls and 3 boys:
\(\binom{3}{2} \cdot \binom{6}{3} = 3 \cdot \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 3 \cdot 20 = 60\) (A1).
Case 2: Exactly 3 girls and 2 boys:
\(\binom{3}{3} \cdot \binom{6}{2} = 1 \cdot \frac{6 \cdot 5}{2 \cdot 1} = 1 \cdot 15 = 15\) (A1).
Total: \( 60 + 15 = 75 \).