| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Selection from categorized items |
| Difficulty | Moderate -0.8 This is a straightforward application of the multiplication principle and basic combinatorics. Part (a)(i) is simple multiplication (3×5×2×3), part (a)(ii) requires counting three cases but is still routine, and part (b) is a standard multinomial coefficient calculation. All techniques are direct applications of fundamental counting principles with no problem-solving insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(3 \times 5 \times 3 \times 2\) or \({}_{3}C_1 \times {}_{5}C_1 \times {}_{3}C_1 \times 2 = 90\) | M1 | For multiplying \(3 \times 5 \times 3\) |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \((3\times5\times2)+(3\times3)+(5\times2\times3) = 69\) | M1 | For summing options that show S&M, S&D, M&D |
| M1 | \(3\times5\times a + 3\times3\times b + 5\times3\times c\) seen for integers \(a,b,c\) | |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \({}_{14}C_5 \times {}_{9}C_5 \times {}_{4}C_4\) or equivalent \(= 252252\) | M1 | For using combinations not all \({}_{14}C\ldots\) |
| M1 | For multiplying choices for two or three groups | |
| A1 | For correct answer; NB \(14!/5!5!4!\) scores M2 and A1 if correct answer |
# Question 5:
## Part (a)(i)
$3 \times 5 \times 3 \times 2$ or ${}_{3}C_1 \times {}_{5}C_1 \times {}_{3}C_1 \times 2 = 90$ | M1 | For multiplying $3 \times 5 \times 3$
| A1 | For correct answer
**Total: 2**
## Part (a)(ii)
$(3\times5\times2)+(3\times3)+(5\times2\times3) = 69$ | M1 | For summing options that show S&M, S&D, M&D
| M1 | $3\times5\times a + 3\times3\times b + 5\times3\times c$ seen for integers $a,b,c$
| A1 | For correct answer
**Total: 3**
## Part (b)
${}_{14}C_5 \times {}_{9}C_5 \times {}_{4}C_4$ or equivalent $= 252252$ | M1 | For using combinations not all ${}_{14}C\ldots$
| M1 | For multiplying choices for two or three groups
| A1 | For correct answer; NB $14!/5!5!4!$ scores M2 and A1 if correct answer
**Total: 3**
---5
\begin{enumerate}[label=(\alph*)]
\item The menu for a meal in a restaurant is as follows.
\begin{displayquote}
Starter Course\\
Melon\\
or\\
Soup\\
or\\
Smoked Salmon
\end{displayquote}
\begin{displayquote}
Main Course\\
Chicken\\
or\\
Steak\\
or\\
Lamb Cutlets\\
or\\
Vegetable Curry\\
or\\
Fish
\end{displayquote}
\begin{displayquote}
Dessert Course\\
Cheesecake\\
or\\
Ice Cream\\
or\\
Apple Pie\\
All the main courses are served with salad and either\\
new potatoes or french fries.
\begin{enumerate}[label=(\roman*)]
\item How many different three-course meals are there?
\item How many different choices are there if customers may choose only two of the three courses?
\end{enumerate}\item In how many ways can a group of 14 people eating at the restaurant be divided between three tables seating 5, 5 and 4?
\end{displayquote}
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2004 Q5 [8]}}