| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with alternating patterns |
| Difficulty | Standard +0.3 This is a multi-part permutations question requiring systematic counting and probability calculations. Part (i)(a) is routine (8!), part (i)(b) requires recognizing alternating arrangements (2×4!×4!), and parts (ii) involve conditional arrangements with adjacency and separation constraints. While it requires careful organization across multiple parts, each individual step uses standard A-level techniques without requiring novel insight or particularly complex reasoning. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8! = 40320\) | M1, A1 2 | Allow \(^4P_4\) & \(^3P_3\) instead of \(3!\) & \(4!\) throughout Q6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{4}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} \times 2\) | M1, M1dep | \(4! \times 4! \div 8! \times 2\); allow \(4! \times 4! + 4! \times 4! \div 8!\); allow 1 above for M1 only |
| \(= \frac{1}{35}\) or \(0.0286\) (3 sfs) | A1 3 | oe, eg \(\frac{1152}{40320}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4! \times 4! = 576\) | M1, A1 2 | allow \(4! \times 4! \times 2\): M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{16}\) or \(0.0625\) | B1 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separated by 5 or 6 questions stated or illustrated | M1 | allow 5 only or 6 only or (4, 5 or 6); can be implied by next M2 or M1 |
| \(\frac{1}{4} \times \frac{1}{4} \times 3\) or \(\frac{1}{16} \times 3\); \((\frac{1}{4} \times \frac{1}{4}\) or \(\frac{1}{16}\) alone or \(\times(2\) or \(6)\): M1) | M2 | \(3! \times 3! \times 3\); \((3!\times 3!\) alone or \(\times(2\) or \(6)\); or \((3! + 3!) \times 3\): M1) \((\div 576)\) |
| \(\frac{3}{16}\) or \(0.1875\) or \(0.188\) | A1 4 | correct ans, but clearly B, J sep by 4: M0M2A0 |
| Alternative: \(1 - P(\text{sep by } 0,1,2,3,(4))\): \(1-(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}x\frac{3}{4}+\frac{1}{4}x\frac{1}{2})\) or \(1-(\frac{1}{4}x\frac{1}{4}+\frac{1}{2}x\frac{1}{4}+\frac{3}{4}x\frac{1}{4}+1x\frac{1}{4}+\frac{3}{4}x\frac{1}{4})\) | M1, M2 | (one omit: M1) |
## Question 6:
**Part (i)(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $8! = 40320$ | M1, A1 2 | Allow $^4P_4$ & $^3P_3$ instead of $3!$ & $4!$ throughout Q6 |
**Part (i)(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{4}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} \times 2$ | M1, M1dep | $4! \times 4! \div 8! \times 2$; allow $4! \times 4! + 4! \times 4! \div 8!$; allow 1 above for M1 only |
| $= \frac{1}{35}$ or $0.0286$ (3 sfs) | A1 3 | oe, eg $\frac{1152}{40320}$ |
**Part (ii)(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $4! \times 4! = 576$ | M1, A1 2 | allow $4! \times 4! \times 2$: M1 |
**Part (ii)(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{16}$ or $0.0625$ | B1 1 | |
**Part (ii)(c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Separated by 5 or 6 questions stated or illustrated | M1 | allow 5 only or 6 only or (4, 5 or 6); can be implied by next M2 or M1 |
| $\frac{1}{4} \times \frac{1}{4} \times 3$ or $\frac{1}{16} \times 3$; $(\frac{1}{4} \times \frac{1}{4}$ or $\frac{1}{16}$ alone or $\times(2$ or $6)$: M1) | M2 | $3! \times 3! \times 3$; $(3!\times 3!$ alone or $\times(2$ or $6)$; or $(3! + 3!) \times 3$: M1) $(\div 576)$ |
| $\frac{3}{16}$ or $0.1875$ or $0.188$ | A1 4 | correct ans, but clearly B, J sep by 4: M0M2A0 |
| Alternative: $1 - P(\text{sep by } 0,1,2,3,(4))$: $1-(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}x\frac{3}{4}+\frac{1}{4}x\frac{1}{2})$ or $1-(\frac{1}{4}x\frac{1}{4}+\frac{1}{2}x\frac{1}{4}+\frac{3}{4}x\frac{1}{4}+1x\frac{1}{4}+\frac{3}{4}x\frac{1}{4})$ | M1, M2 | (one omit: M1) |
---6 A test consists of 4 algebra questions, A, B, C and D, and 4 geometry questions, G, H, I and J.\\
The examiner plans to arrange all 8 questions in a random order, regardless of topic.
\begin{enumerate}[label=(\roman*)]
\item (a) How many different arrangements are possible?\\
(b) Find the probability that no two Algebra questions are next to each other and no two Geometry questions are next to each other.
Later, the examiner decides that the questions should be arranged in two sections, Algebra followed by Geometry, with the questions in each section arranged in a random order.
\item (a) How many different arrangements are possible?\\
(b) Find the probability that questions A and H are next to each other.\\
(c) Find the probability that questions B and J are separated by more than four other questions.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2009 Q6 [12]}}