| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with adjacency requirements |
| Difficulty | Moderate -0.3 This is a standard permutations question with familiar patterns: arrangements with repeated letters, treating groups as single units, and converting combinations to permutations. Part (a) uses the standard formula for repeated letters (12!/(2!2!2!)), part (b)(i) is straightforward multiplication by 4!, and part (b)(ii) combines grouping with internal arrangements (4! × (3!)^4). All techniques are routine textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{12!}{2!3!2!} = 19958400\ (20{,}000{,}000)\) | M1 | Dividing by \(2!\ 3!\ 2!\) |
| A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4!}{2!} \times \frac{9!}{2!3!} = 362880\) | B1 | \(4!\) seen multiplied |
| B1 | \(9!\) or \(9 \times 8!\) seen multiplied | |
| B1 [3] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3876 \times 4!\) | M1 | Multiplying by \(4!\) |
| \(= 93024\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((3!)^4 \times 4!\) | M1 | \(3!\) or \(6\) or \(4!\) seen |
| \(= 31104\) | A1 [2] | Correct final answer |
## Question 6:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{12!}{2!3!2!} = 19958400\ (20{,}000{,}000)$ | M1 | Dividing by $2!\ 3!\ 2!$ |
| | A1 [2] | Correct answer |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4!}{2!} \times \frac{9!}{2!3!} = 362880$ | B1 | $4!$ seen multiplied |
| | B1 | $9!$ or $9 \times 8!$ seen multiplied |
| | B1 [3] | Correct final answer |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3876 \times 4!$ | M1 | Multiplying by $4!$ |
| $= 93024$ | A1 [2] | Correct answer |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3!)^4 \times 4!$ | M1 | $3!$ or $6$ or $4!$ seen |
| $= 31104$ | A1 [2] | Correct final answer |
---6
\begin{enumerate}[label=(\alph*)]
\item Find the number of different ways in which the 12 letters of the word STRAWBERRIES can be arranged
\begin{enumerate}[label=(\roman*)]
\item if there are no restrictions,
\item if the 4 vowels $\mathrm { A } , \mathrm { E } , \mathrm { E } , \mathrm { I }$ must all be together.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item 4 astronauts are chosen from a certain number of candidates. If order of choosing is not taken into account, the number of ways the astronauts can be chosen is 3876 . How many ways are there if order of choosing is taken into account?
\item 4 astronauts are chosen to go on a mission. Each of these astronauts can take 3 personal possessions with him. How many different ways can these 12 possessions be arranged in a row if each astronaut's possessions are kept together?
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2011 Q6 [9]}}