| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Combinatorial selection with category constraints |
| Difficulty | Moderate -0.8 This is a straightforward combinations question requiring basic counting principles. Part (i) uses simple nCr calculations, part (ii) is elementary probability, and part (iii) requires complementary counting (subtract restricted cases). All techniques are standard S1 material with no novel problem-solving required, making it easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Rice dishes | Main dishes | Vegetable dishes |
| Boiled rice | Chicken | Mushrooms |
| Fried rice | Beef | Cauliflower |
| Pilau rice | Lamb | Spinach |
| Keema rice | Mixed grill | Lentils |
| Prawn | Potatoes | |
| Vegetarian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(^6C_2×^6C_1×^6C_4\) or \(6×20×5\) = 600 | M1M1, A1 3 | M1 for any 2 correct combs seen, even if added |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\frac{2}{4}\) or \(\frac{^3C_1}{^4C_2}\) or \(\frac{3C_1^6C_5^6C_4}{oe}\) or \(\frac{^3C_0^6C_5^6C_4}{600}\) = \(\frac{1}{2}\) oe | M1, A1 2 | or \(\frac{1}{4}×1+\frac{3}{4}×\frac{1}{3}\) or \(\frac{1}{4}×2\) or \(\frac{1}{4}+\frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(^3C_1×^6C_3(×^3C_4)+^3C_2×^6C_3(×^3C_4)\) or 360 | M1M1, A1 3 | M1 either product seen, even if × or ÷ by something |
## Part i:
**Answer/Working:** $^6C_2×^6C_1×^6C_4$ or $6×20×5$ = 600 | **M1M1, A1 3** | M1 for any 2 correct combs seen, even if added
## Part ii:
**Answer/Working:** $\frac{2}{4}$ or $\frac{^3C_1}{^4C_2}$ or $\frac{3C_1^6C_5^6C_4}{oe}$ or $\frac{^3C_0^6C_5^6C_4}{600}$ = $\frac{1}{2}$ oe | **M1, A1 2** | or $\frac{1}{4}×1+\frac{3}{4}×\frac{1}{3}$ or $\frac{1}{4}×2$ or $\frac{1}{4}+\frac{1}{4}$
## Part iii:
**Answer/Working:** $^3C_1×^6C_3(×^3C_4)+^3C_2×^6C_3(×^3C_4)$ or 360 | **M1M1, A1 3** | M1 either product seen, even if × or ÷ by something
**Total: 8 marks**
---The menu below shows all the dishes available at a certain restaurant.
\begin{center}
\begin{tabular}{|c|c|c|}
\hline
Rice dishes & Main dishes & Vegetable dishes \\
\hline
Boiled rice & Chicken & Mushrooms \\
Fried rice & Beef & Cauliflower \\
Pilau rice & Lamb & Spinach \\
Keema rice & Mixed grill & Lentils \\
& Prawn & Potatoes \\
& Vegetarian & \\
\hline
\end{tabular}
\end{center}
A group of friends decide that they will share a total of 2 different rice dishes, 3 different main dishes and 4 different vegetable dishes from this menu. Given these restrictions,
\begin{enumerate}[label=(\roman*)]
\item find the number of possible combinations of dishes that they can choose to share, [3]
\item assuming that all choices are equally likely, find the probability that they choose boiled rice. [2]
\end{enumerate}
The friends decide to add a further restriction as follows. If they choose boiled rice, they will not choose potatoes.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the number of possible combinations of dishes that they can now choose. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2010 Q7 [8]}}