Solution — 2020 Question 8 Permutations & Combinations

(a)

\begin{matrix} Number of ways \\ = (\binom{12}{5}) \\ = 792 ∎ \end{matrix}

(b)

For more girls than boys in a team of 5, we need at least 3 girls. We consider three cases.

Case 1: 3 girls, 2 boys

\begin{matrix} Number of ways \\ = (\binom{7}{3}) \times (\binom{5}{2}) \\ = 350 \end{matrix}

Case 2: 4 girls, 1 boy

\begin{matrix} Number of ways \\ = (\binom{7}{4}) \times (\binom{5}{1}) \\ = 175 \end{matrix}

Case 3: 5 girls, 0 boys

\begin{matrix} Number of ways \\ = (\binom{7}{5}) \times (\binom{5}{0}) \\ = 21 \end{matrix}

\begin{matrix} Total number of ways \\ = 350 + 175 + 21 \\ = 546 ∎ \end{matrix}

(c)

Treat the 2 boys as a single block. There are then 4 units (the block and the 3 girls) to arrange in a line, and the 2 boys within the block can be arranged among themselves.

\begin{matrix} Number of ways \\ = 4! \times 2! \\ = 48 ∎ \end{matrix}