| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Combinatorial selection with category constraints |
| Difficulty | Easy -1.2 This is a straightforward permutations question testing basic counting principles. Part (i) is a direct application of ordered selection without replacement (20×19×18), while part (ii) requires minimal adjustment by subtracting cases where one student wins all three prizes. Both parts are routine calculations with no conceptual difficulty beyond recognizing the counting structure. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(20 \times 19 \times 18 = 6840\) | M1, A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(20^3 - 20 = 7980\) | M1 for figures \(-20\), A1 | [2] |
| TOTAL | [4] |
## (i)
$20 \times 19 \times 18 = 6840$ | M1, A1 | [2]
## (ii)
$20^3 - 20 = 7980$ | M1 for figures $-20$, A1 | [2]
**TOTAL** | | [4]
---Three prizes, one for English, one for French and one for Spanish, are to be awarded in a class of 20 students.
Find the number of different ways in which the three prizes can be awarded if
\begin{enumerate}[label=(\roman*)]
\item no student may win more than 1 prize, [2]
\item no student may win all 3 prizes. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2010 Q6 [4]}}