| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with adjacency requirements |
| Difficulty | Moderate -0.3 This is a standard permutations question with restrictions. Part (i) uses the classic 'treat as a block' technique (7! × 6!), and part (ii) requires arranging boys first then placing girls in gaps (6! × 7P4). Both are textbook methods with straightforward application, making this slightly easier than average but still requiring proper technique. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(6! \times 5! = 86400\) | B1, B1, B1 3 | \(6!\) or seen multiplied by integer \(k \geq 1\) or \(5!\) or seen multiplied by integer \(k \geq 1\) or Correct final answer |
| (ii) \(6! \times 7 \times 6 \times 5 \times 4 = 604800\) | B1, B1, B1 3 | \(6!\) seen mult by integer \(k \geq 1\) or Mult by \(^7P_4\) or Correct final answer |
**(i)** $6! \times 5! = 86400$ | B1, B1, B1 3 | $6!$ or seen multiplied by integer $k \geq 1$ or $5!$ or seen multiplied by integer $k \geq 1$ or Correct final answer
**(ii)** $6! \times 7 \times 6 \times 5 \times 4 = 604800$ | B1, B1, B1 3 | $6!$ seen mult by integer $k \geq 1$ or Mult by $^7P_4$ or Correct final answer
---Find the number of different ways that 6 boys and 4 girls can stand in a line if
\begin{enumerate}[label=(\roman*)]
\item all 6 boys stand next to each other, [3]
\item no girl stands next to another girl. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2014 Q2 [6]}}