| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with couples/pairs |
| Difficulty | Standard +0.3 This is a standard permutations question with couples constraints. Part (a)(i) uses the classic 'treat couples as units' approach (6! × 2^6), part (a)(ii) requires careful counting with fixed positions, and part (b) involves complementary counting or direct enumeration of grouping cases. All techniques are routine for A-level statistics students, requiring no novel insight beyond textbook methods. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(6!\times2^6\) | M1 | \(6!\times a\), \(a\) integer \(>1\) |
| M1 | \(b\times2^6\), \(b\) integer \(\geq1\) | |
| \(= 46080\) | A1 | Accurate answer required. SC B1 for 46080 if M0 M0 www |
| Answer | Marks | Guidance |
|---|---|---|
| \(12\times10\times8\times6\times4\times2\) | M1 | \(c\times d\times e\times f\times g\times h\); \(2\leq c,d,e,f,g,h\) (different integers) \(\leq12\) |
| M1 | Correct unsimplified | |
| \(= 46080\) | A1 | Accurate answer required. SC B1 for 46080 if M0 M0 www |
| Answer | Marks | Guidance |
|---|---|---|
| \(5!\times5!\times2\times2\) | M1 | \(5!\times5!\times k\), \(k\) positive integer, 1 may be implied (no adding/subtracting) |
| \(= 57600\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| In the group of 5: \(\frac{5}{12}\times\frac{4}{11} = \left[\frac{20}{132}, \frac{5}{33}\right]\) | B1 | Correct probability for one identified scenario |
| In the group of 4: \(\frac{4}{12}\times\frac{3}{11} = \left[\frac{12}{132}, \frac{1}{11}\right]\) | M1 | Denominator \(12\times11\) for all probabilities (1, 2 or 3 scenarios) |
| In the group of 3: \(\frac{3}{12}\times\frac{2}{11} = \left[\frac{6}{132}, \frac{1}{22}\right]\) | A1 | 3 correct probabilities, accept unsimplified |
| \(\frac{5}{12}\times\frac{4}{11}+\frac{4}{12}\times\frac{3}{11}+\frac{3}{12}\times\frac{2}{11}\) | M1 | Adding probabilities for 3 correct scenarios |
| \(\frac{19}{66},\ 0.288\) | A1 | 0.2878787 to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| In the group of 5: \(^{10}C_3 \times ^7C_4 = 120 \times 35 = 4200\) | B1 | Correct value of one identified scenario seen, accept unsimplified |
| In the group of 4: \(^{10}C_2 \times ^8C_5 = 45 \times 56 = 2520\) | M1 | \(^{12}C_a \times ^{12-a}C_b\), \(a = 3, 4, 5\); \(b = 3, 4, 5\) \((a \neq b)\) |
| In the group of 3: \(^{10}C_1 \times ^9C_5 = 10 \times 126 = 1260\) | ||
| Total arrangements of 3 groups: \(^{12}C_5 \times ^7C_4 = 792 \times 35 = 27720\), or \(^{12}C_3 \times ^9C_4\) or \(^{12}C_4 \times ^8C_5\) | A1 | 27720 seen alone or as denominator of probability – accept unsimplified. SC B1 if M0 |
| \(4200 + 2520 + 1260 = 7980\) | M1 | Values of 3 correct scenarios added, accept unsimplified – or correct |
| Probability \(= \dfrac{7980}{27720} = \dfrac{19}{66} \approx 0.288\) | A1 | 0.2878787 to at least 3SF |
## Question 6(a)(i):
**Method 1:**
$6!\times2^6$ | M1 | $6!\times a$, $a$ integer $>1$
| M1 | $b\times2^6$, $b$ integer $\geq1$
$= 46080$ | A1 | Accurate answer required. **SC B1** for 46080 if M0 M0 www
**Alternative Method:**
$12\times10\times8\times6\times4\times2$ | M1 | $c\times d\times e\times f\times g\times h$; $2\leq c,d,e,f,g,h$ (different integers) $\leq12$
| M1 | Correct unsimplified
$= 46080$ | A1 | Accurate answer required. **SC B1** for 46080 if M0 M0 www
---
## Question 6(a)(ii):
$5!\times5!\times2\times2$ | M1 | $5!\times5!\times k$, $k$ positive integer, 1 may be implied (no adding/subtracting)
$= 57600$ | A1 |
---
## Question 6(b):
In the group of 5: $\frac{5}{12}\times\frac{4}{11} = \left[\frac{20}{132}, \frac{5}{33}\right]$ | B1 | Correct probability for one identified scenario
In the group of 4: $\frac{4}{12}\times\frac{3}{11} = \left[\frac{12}{132}, \frac{1}{11}\right]$ | M1 | Denominator $12\times11$ for all probabilities (1, 2 or 3 scenarios)
In the group of 3: $\frac{3}{12}\times\frac{2}{11} = \left[\frac{6}{132}, \frac{1}{22}\right]$ | A1 | 3 correct probabilities, accept unsimplified
$\frac{5}{12}\times\frac{4}{11}+\frac{4}{12}\times\frac{3}{11}+\frac{3}{12}\times\frac{2}{11}$ | M1 | Adding probabilities for 3 correct scenarios
$\frac{19}{66},\ 0.288$ | A1 | 0.2878787 to at least 3SF
## Question 6(b):
**Method 2 – number of arrangements of J & K being placed:**
| Answer | Mark | Guidance |
|--------|------|----------|
| In the group of 5: $^{10}C_3 \times ^7C_4 = 120 \times 35 = 4200$ | **B1** | Correct value of one identified scenario seen, accept unsimplified |
| In the group of 4: $^{10}C_2 \times ^8C_5 = 45 \times 56 = 2520$ | **M1** | $^{12}C_a \times ^{12-a}C_b$, $a = 3, 4, 5$; $b = 3, 4, 5$ $(a \neq b)$ |
| In the group of 3: $^{10}C_1 \times ^9C_5 = 10 \times 126 = 1260$ | | |
| Total arrangements of 3 groups: $^{12}C_5 \times ^7C_4 = 792 \times 35 = 27720$, or $^{12}C_3 \times ^9C_4$ or $^{12}C_4 \times ^8C_5$ | **A1** | 27720 seen alone or as denominator of probability – accept unsimplified. **SC B1** if M0 |
| $4200 + 2520 + 1260 = 7980$ | **M1** | Values of 3 correct scenarios added, accept unsimplified – or correct |
| Probability $= \dfrac{7980}{27720} = \dfrac{19}{66} \approx 0.288$ | **A1** | 0.2878787 to at least 3SF |
**Note:** Alternative arrangement calculations possible e.g.:
- In the group of 5: $^{10}C_3 \times ^7C_4 = 120 \times 35 = 4200$
- In the group of 4: $^{10}C_5 \times ^5C_2 = 252 \times 10 = 2520$
- In the group of 3: $^{10}C_5 \times ^5C_4 = 252 \times 5 = 1260$
**Total marks: 5**6 Jai and his wife Kaz are having a party. Jai has invited five friends and each friend will bring his wife.
\begin{enumerate}[label=(\alph*)]
\item At the beginning of the party, the 12 people will stand in a line for a photograph.
\begin{enumerate}[label=(\roman*)]
\item How many different arrangements are there of the 12 people if Jai stands next to Kaz and each friend stands next to his own wife?
\item How many different arrangements are there of the 12 people if Jai and Kaz occupy the two middle positions in the line, with Jai's five friends on one side and the five wives of the friends on the other side?
\end{enumerate}\item For a competition during the party, the 12 people are divided at random into a group of 5, a group of 4 and a group of 3 .
Find the probability that Jai and Kaz are in the same group as each other.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q6 [10]}}