Solution — 2018 Question 7 Permutations & Combinations

(a)

\begin{matrix} Number of selections \\ = (\binom{8}{3}) \times (\binom{4}{1}) \times (\binom{6}{2}) \\ = 3360 ∎ \end{matrix}

(b)

Case 1: 0 family guitarists chosen

\begin{matrix} Number of ways \\ = (\binom{3}{0}) \times (\binom{5}{3}) \\ = 10 \end{matrix}

Case 2: Exactly 1 family guitarist chosen

\begin{matrix} Number of ways \\ = (\binom{3}{1}) \times (\binom{5}{2}) \\ = 30 \end{matrix}

\begin{matrix} Required number of selections \\ = 10 + 30 \\ = 40 ∎ \end{matrix}

(c)

\begin{matrix} P (no family guitarists in band) \\ = \frac{(\binom{5}{3})}{(\binom{8}{3})} \\ = \frac{5}{28} ∎ \end{matrix}