Solution — 2017 Question 8 Permutations & Combinations

(a)

\begin{matrix} Number of codes \\ = (\binom{6}{3}) \times 3! \times (\binom{8}{3}) \times 3! \\ = 40320 ∎ \end{matrix}

(b)

With repetitions allowed, total number of codes $= 6^{3} \times 8^{3}$

(i)

For the digit $5$ to appear exactly once, one of the choices must be 5 (1 way) and we fill the remaining 2 digit positions with any of the 5 other digits. There are 3 ways to rearrange these digits (5XX, X5X, XX5)

For the letter $H$ to appear exactly twice, H must appear twice (1 way) and we fill the remaining letter position with any of the 7 other letters. There are also 3 ways to rearrange these letters (HHX, HXH, XHH)

\begin{matrix} P (5 exactly once and H exactly twice) \\ = \frac{(3 \times 5^{2}) \times (3 \times 7)}{6^{3} \times 8^{3}} \\ = \frac{175}{12 288} ∎ \\ = 0.0142 (to 3 s.f.) ∎ \end{matrix}

(ii)

Case 1: 2 is the first character but H is not the sixth character

\begin{matrix} Number of ways \\ = 1 \times 6^{2} \times 8^{2} \times 7 \\ = 16128 \end{matrix}

Case 2: H is the sixth character but 2 is not the first character

\begin{matrix} Number of ways \\ = 1 \times 6^{2} \times 8^{2} \times 7 \\ = 11520 \end{matrix}

\begin{matrix} P (required) \\ = \frac{16128 + 11520}{6^{3} \times 8^{3}} \\ = \frac{1}{4} ∎ \end{matrix}