| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with grouped categories |
| Difficulty | Moderate -0.8 Part (i) is a standard grouped arrangement problem (treat boys/girls as blocks: 2! × 5! × 6!). Part (ii) requires the insight to place girls first then insert boys in gaps, but this is a well-known textbook technique. Both parts involve straightforward application of permutation formulas with minimal problem-solving demand. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5! \times 6! \times 2\) | B1 | \(k \times 5!\) or \(m \times 6!\) (\(k,m\) integer, \(k,m \geq 1\)), no inappropriate addition |
| B1 | \(n \times 5! \times 6!\) (\(n\) integer, \(n \geq 1\)), no inappropriate addition | |
| \(= 172800\) | B1 | Correct final answer, isw rounding (www scores B3). All marks based on their final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ldots G \ldots G \ldots G \ldots G \ldots G \ldots G \ldots\); No. ways girls placed \(\times\) No. ways boys placed in gaps | M1 | \(k \times 6!\) or \(k \times {}^7P_5\) (\(k\) is an integer, \(k \geq 1\)) no inappropriate add. (\({}^7P_5 \equiv 7 \times 6 \times 5 \times 4 \times 3\) or \({}^7C_5 \times 5!\)) |
| \(6! \times {}^7P_5\) | M1 | Correct unsimplified expression |
| \(= 1814400\) | A1 | Correct exact final answer (ignore subsequent rounding) |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5! \times 6! \times 2$ | **B1** | $k \times 5!$ or $m \times 6!$ ($k,m$ integer, $k,m \geq 1$), no inappropriate addition |
| | **B1** | $n \times 5! \times 6!$ ($n$ integer, $n \geq 1$), no inappropriate addition |
| $= 172800$ | **B1** | Correct final answer, isw rounding (www scores B3). All marks based on their **final** answer |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ldots G \ldots G \ldots G \ldots G \ldots G \ldots G \ldots$; No. ways girls placed $\times$ No. ways boys placed in gaps | **M1** | $k \times 6!$ or $k \times {}^7P_5$ ($k$ is an integer, $k \geq 1$) no inappropriate add. (${}^7P_5 \equiv 7 \times 6 \times 5 \times 4 \times 3$ or ${}^7C_5 \times 5!$) |
| $6! \times {}^7P_5$ | **M1** | Correct unsimplified expression |
| $= 1814400$ | **A1** | Correct exact final answer (ignore subsequent rounding) |4 (i) Find the number of different ways that 5 boys and 6 girls can stand in a row if all the boys stand together and all the girls stand together.\\
(ii) Find the number of different ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy.\\
\hfill \mbox{\textit{CAIE S1 2018 Q4 [6]}}