| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Standard +0.3 This is a standard permutations question with repeated letters and basic constraints. Part (a) is routine application of n!/r! formula, part (b) requires complementary counting with fixed positions (a common textbook technique), and part (c) is straightforward probability with grouping. All parts use well-practiced methods without requiring novel insight, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks |
|---|---|
| \(\left[\dfrac{9!}{2!3!} =\right] 30240\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{7!}{3!} - 5! =\) | B1 | \(\dfrac{7!}{3!} - e\), \(^7P_4 - e\), \(e\) a positive integer |
| M1 | \(f - \dfrac{5!}{r!}\), \(f > 120\), \(r = 1, 2\) | |
| \(720\) | A1 | If no marks scored SC B1 for \(840 - 120 = 720\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(^5C_3 \times 4! + ^4C_1 \times 5!\) or \(\dfrac{^5P_3}{3!} \times 4! + ^5P_2 \times 4!\) | (B1) | One of \(^5C_3 \times 4!\), \(\dfrac{^5P_3}{3!} \times 4!\), \(^4C_1 \times 5!\) or \(^5P_2 \times 4!\) seen |
| (M1) | \(a \times 4! + b \times 5!\) where \(a\) and \(b\) are integers between 1 and 10 inclusive, or \(c \times 4! + d \times 4!\) where \(c\) and \(d\) are integers between 1 and 20 inclusive | |
| \(240 + 480 = 720\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Group of 5: \(\quad 3\text{Rs}\ 2\text{Es} = 1\); \(\quad 3\text{Rs}\ 1\text{E} = ^2C_1 \times ^4C_1 = 8\); \(\quad 3\text{Rs}\ 0\text{Es} = ^4C_2 = 6\) Group of 4: \(\quad 3\text{Rs}\ 1\text{E} = ^2C_1 = 2\); \(\quad 3\text{Rs}\ 0\text{Es} = ^4C_1 = 4\) | B1 | Correct no. of ways for two correct identified scenarios other than three Rs two Es |
| \([\text{Total} =]\ 21\) | M1 | No. of ways for five correct identified scenarios added or correct |
| \([\text{Number of ways of splitting into two groups} =]\ ^9C_5 (= 126)\) seen as a denominator | M1 | Accept evaluated, accept \(^9C_4\) |
| Probability \(= \dfrac{21}{126} = \dfrac{1}{6}\ (0.167)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\text{Rs in Group of 5} = ^6C_2 = 15\); \(\quad 3\text{Rs in Group of 4} = ^6C_1 = 6\) | (B1) | One correct case evaluated accurately and linked with correct scenario |
| \([\text{Total} =]\ 21\) | (M1) | No. of ways for two correct scenarios added or correct |
| \(^9C_5 (= 126)\) seen as a denominator | (M1) | Accept evaluated, accept \(^9C_4\) |
| Probability \(= \dfrac{21}{126} = \dfrac{1}{6}\ (0.167)\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\text{Rs in Group of 5}: \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{3}{7} = \dfrac{15}{126}\) | (B1) | For one correct product unsimplified and linked with correct scenario |
| \(3\text{Rs in Group of 4}: \dfrac{4}{9} \times \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{126}\) | (M1) | For second correct product |
| \(\dfrac{15}{126} + \dfrac{6}{126}\) | (M1) | For adding probabilities of two correct scenarios or correct |
| Probability \(= \dfrac{21}{126} = \dfrac{1}{6}\ (0.167)\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Two correct probabilities linked with correct scenarios | (B1) | Two correct probabilities linked with correct scenarios, accept unsimplified. |
| Four probabilities with denominators including factor of \(9 \times 8 \times 7 \times 6 \times n\) | (M1) | Four probabilities with denominators including a factor of \(9 \times 8 \times 7 \times 6 \times n\), where \(n\) is 1 or 5. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{1+8+6+6}{126}\) | (M1) | Probabilities of four correct scenarios added or correct. |
| Probability \(= \dfrac{21}{126} = \dfrac{1}{6} \ (0.167)\) | (A1) | |
| 4 |
## Question 6(a):
| $\left[\dfrac{9!}{2!3!} =\right] 30240$ | B1 | |
**Total: 1 mark**
---
## Question 6(b):
**Method 1:**
| $\dfrac{7!}{3!} - 5! =$ | B1 | $\dfrac{7!}{3!} - e$, $^7P_4 - e$, $e$ a positive integer |
|---|---|---|
| | M1 | $f - \dfrac{5!}{r!}$, $f > 120$, $r = 1, 2$ |
| $720$ | A1 | If no marks scored SC B1 for $840 - 120 = 720$ |
**Method 2:**
| $^5C_3 \times 4! + ^4C_1 \times 5!$ or $\dfrac{^5P_3}{3!} \times 4! + ^5P_2 \times 4!$ | (B1) | One of $^5C_3 \times 4!$, $\dfrac{^5P_3}{3!} \times 4!$, $^4C_1 \times 5!$ or $^5P_2 \times 4!$ seen |
|---|---|---|
| | (M1) | $a \times 4! + b \times 5!$ where $a$ and $b$ are integers between 1 and 10 inclusive, or $c \times 4! + d \times 4!$ where $c$ and $d$ are integers between 1 and 20 inclusive |
| $240 + 480 = 720$ | (A1) | |
**Total: 3 marks**
---
## Question 6(c):
**Method 1:**
| Group of 5: $\quad 3\text{Rs}\ 2\text{Es} = 1$; $\quad 3\text{Rs}\ 1\text{E} = ^2C_1 \times ^4C_1 = 8$; $\quad 3\text{Rs}\ 0\text{Es} = ^4C_2 = 6$ Group of 4: $\quad 3\text{Rs}\ 1\text{E} = ^2C_1 = 2$; $\quad 3\text{Rs}\ 0\text{Es} = ^4C_1 = 4$ | B1 | Correct no. of ways for two correct identified scenarios other than three Rs two Es |
|---|---|---|
| $[\text{Total} =]\ 21$ | M1 | No. of ways for five correct identified scenarios added or correct |
| $[\text{Number of ways of splitting into two groups} =]\ ^9C_5 (= 126)$ seen as a denominator | M1 | Accept evaluated, accept $^9C_4$ |
| Probability $= \dfrac{21}{126} = \dfrac{1}{6}\ (0.167)$ | A1 | |
**Method 2:**
| $3\text{Rs in Group of 5} = ^6C_2 = 15$; $\quad 3\text{Rs in Group of 4} = ^6C_1 = 6$ | (B1) | One correct case evaluated accurately and linked with correct scenario |
|---|---|---|
| $[\text{Total} =]\ 21$ | (M1) | No. of ways for two correct scenarios added or correct |
| $^9C_5 (= 126)$ seen as a denominator | (M1) | Accept evaluated, accept $^9C_4$ |
| Probability $= \dfrac{21}{126} = \dfrac{1}{6}\ (0.167)$ | (A1) | |
**Method 3:**
| $3\text{Rs in Group of 5}: \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{3}{7} = \dfrac{15}{126}$ | (B1) | For one correct product unsimplified and linked with correct scenario |
|---|---|---|
| $3\text{Rs in Group of 4}: \dfrac{4}{9} \times \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{126}$ | (M1) | For second correct product |
| $\dfrac{15}{126} + \dfrac{6}{126}$ | (M1) | For adding probabilities of two correct scenarios or correct |
| Probability $= \dfrac{21}{126} = \dfrac{1}{6}\ (0.167)$ | (A1) | |
**Total: 4 marks**
## Question 6(c): Method 4 - Probability Method
| Answer | Marks | Guidance |
|--------|-------|----------|
| Two correct probabilities linked with correct scenarios | **(B1)** | Two correct probabilities linked with correct scenarios, accept unsimplified. |
| Four probabilities with denominators including factor of $9 \times 8 \times 7 \times 6 \times n$ | **(M1)** | Four probabilities with denominators including a factor of $9 \times 8 \times 7 \times 6 \times n$, where $n$ is 1 or 5. |
**Group of 5:**
$$\text{2Es:} \quad \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7} \times \frac{2}{6} \times \frac{1}{5} \times \frac{5!}{3!2!} = \frac{1}{126}$$
$$\text{1E:} \quad \frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} \times \frac{2}{5} \times \frac{5!}{3!} = \frac{8}{126}$$
$$\text{0E:} \quad \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7} \times \frac{4}{6} \times \frac{3}{5} \times \frac{5!}{3!2!} = \frac{6}{126}$$
**Group of 4:**
$$\frac{3}{9} \times \frac{2}{8} \times \frac{1}{7} \times \frac{6}{6} \times \frac{4!}{3!} = \frac{6}{126}$$
| $\dfrac{1+8+6+6}{126}$ | **(M1)** | Probabilities of four correct scenarios added or correct. |
|--------|-------|----------|
| Probability $= \dfrac{21}{126} = \dfrac{1}{6} \ (0.167)$ | **(A1)** | |
| | **4** | |6
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements are there of the 9 letters in the word RECORDERS?
\item How many different arrangements are there of the 9 letters in the word RECORDERS in which there is an E at the beginning, an E at the end and the three Rs are not all together?\\
\includegraphics[max width=\textwidth, alt={}, center]{9b21cc0f-b043-4251-8aa9-cb1e5c2fb5d0-12_2725_40_136_2007}
\begin{center}
\end{center}
The 9 letters of the word RECORDERS are divided at random into two groups: a group of 5 letters and a group of 4 letters.
\item Find the probability that the three Rs are in the same group.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\includegraphics[max width=\textwidth, alt={}, center]{9b21cc0f-b043-4251-8aa9-cb1e5c2fb5d0-14_2715_35_143_2012}
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q6 [8]}}