| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Basic arrangements with repeated letters |
| Difficulty | Moderate -0.8 This is a straightforward permutations question testing basic counting principles. Part (i) is standard permutations with repetition (5!/2!), part (ii) uses the standard 'treat as one unit' technique, and part (iii) requires simple probability calculation. All methods are routine textbook exercises requiring minimal problem-solving insight, making this easier than average for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{5!}{2} = 60\) | M1 A1 2 | Allow 5P3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(4!\) | M1 A1 2 | Allow 2×4! |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{24}{7_5} \times 7_4 or3/5×2/4\) | M1 | allow M1 for \(\frac{7}{5} \times \frac{7}{5}\) ×2 or \(\frac{7_75}{25}\) |
| \(\times 2 = \frac{7}{5}\) oe | M1 M1 A1 3 | or (6×3)÷(i) M2 or 3!÷(i),(6+i)or 6×6 or 36 or 1-correct answer M1 (k,integer ≤5) |
### (i)
$\frac{5!}{2} = 60$ | M1 A1 2 | Allow 5P3
### (ii)
$4!$ | M1 A1 2 | Allow 2×4!
### (iii)
$\frac{24}{7_5} \times 7_4 or3/5×2/4$ | M1 | allow M1 for $\frac{7}{5} \times \frac{7}{5}$ ×2 or $\frac{7_75}{25}$
$\times 2 = \frac{7}{5}$ oe | M1 M1 A1 3 | or (6×3)÷(i) M2 or 3!÷(i),(6+i)or 6×6 or 36 or 1-correct answer M1 (k,integer ≤5)
**Total [7]**
---The five letters of the word NEVER are arranged in random order in a straight line.
\begin{enumerate}[label=(\roman*)]
\item How many different orders of the letters are possible? [2]
\item In how many of the possible orders are the two Es next to each other? [2]
\item Find the probability that the first two letters in the order include exactly one letter E. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2010 Q8 [7]}}