| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | People arrangements in lines |
| Difficulty | Moderate -0.3 This is a standard permutations question with straightforward applications of factorial counting and complementary counting. Part (a) fixes one position (8! arrangements), part (b) uses total minus adjacent (9! - 2×8!), part (c) is a basic combination C(9,5), and part (d) applies simple probability reasoning. All techniques are routine for S1 level with no novel insight required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([8! =]\ 40\,320\) | B1 | Evaluated, exact value only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \(7! \times {}^8C_2 \times 2\) | M1 | \(7! \times k\) seen, \(k\) an integer \(> 1\) |
| M1 | \(m \times n(n-1)\) or \(m \times {}^nC_2\) or \(m \times {}^nP_2\), \(n = 7, 8\) or \(9\), \(m\) an integer \(> 1\) | |
| \(282\,240\) | A1 | Exact value only. SC B1 for final answer 282 240 WWW |
| Method 2: \(9! - 8! \times 2\) | M1 | \(9! - k\), \(k\) an integer \(< 362\,880\) |
| M1 | \(m - 8! \times n\), \(m\) an integer \(> 40\,320\), \(n = 1, 2\) | |
| \(282\,240\) | A1 | Exact value only. SC B1 for final answer 282 240 WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \({}^9C_5 \left[\times {}^4C_4\right]\) | M1 | \({}^9C_x \left[\times {}^{9-x}C_{9-x}\right]\) \(x = 4, 5\). Condone \(\times 1\) for \({}^{9-x}C_{9-x}\). Condone use of P |
| \(126\) | A1 | WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Number of ways with Raman and Sanjay together on back row \(=\)] \({}^7C_3\); [Number of ways with Raman and Sanjay together on front row \(=\)] \({}^7C_2\) | M1 | \({}^7C_x\) seen, \(x = 3\) or \(2\) |
| \([\text{Total} =]\ 35 + 21\) | M1 | Summing two correct scenarios |
| \(56\) | A1 | Evaluated – may be seen used in probability. If M0 scored, SC B1 for 56 WWW |
| \(\text{Probability} = \frac{\textit{their } 56}{\textit{their}(c)} = \frac{56}{126}, \frac{4}{9}, 0.444\) | B1 FT | FT *their* 56 from adding 2 or more scenarios in numerator and *their* (c) or correct as denominator |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[8! =]\ 40\,320$ | B1 | Evaluated, exact value only |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $7! \times {}^8C_2 \times 2$ | M1 | $7! \times k$ seen, $k$ an integer $> 1$ |
| | M1 | $m \times n(n-1)$ or $m \times {}^nC_2$ or $m \times {}^nP_2$, $n = 7, 8$ or $9$, $m$ an integer $> 1$ |
| $282\,240$ | A1 | Exact value only. **SC B1** for final answer 282 240 WWW |
| **Method 2:** $9! - 8! \times 2$ | M1 | $9! - k$, $k$ an integer $< 362\,880$ |
| | M1 | $m - 8! \times n$, $m$ an integer $> 40\,320$, $n = 1, 2$ |
| $282\,240$ | A1 | Exact value only. **SC B1** for final answer 282 240 WWW |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| ${}^9C_5 \left[\times {}^4C_4\right]$ | M1 | ${}^9C_x \left[\times {}^{9-x}C_{9-x}\right]$ $x = 4, 5$. Condone $\times 1$ for ${}^{9-x}C_{9-x}$. Condone use of P |
| $126$ | A1 | WWW |
---
## Question 5(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Number of ways with Raman and Sanjay together on back row $=$] ${}^7C_3$; [Number of ways with Raman and Sanjay together on front row $=$] ${}^7C_2$ | M1 | ${}^7C_x$ seen, $x = 3$ or $2$ |
| $[\text{Total} =]\ 35 + 21$ | M1 | Summing two correct scenarios |
| $56$ | A1 | Evaluated – may be seen used in probability. If **M0** scored, **SC B1** for 56 WWW |
| $\text{Probability} = \frac{\textit{their } 56}{\textit{their}(c)} = \frac{56}{126}, \frac{4}{9}, 0.444$ | B1 FT | FT *their* 56 from adding 2 or more scenarios in numerator and *their* **(c)** or correct as denominator |
---5 Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken.
For the first photograph, the 9 members will stand in a line.
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements of the 9 members are possible in which Raman will be at the centre of the line?
\item How many different arrangements of the 9 members are possible in which Raman and Sanjay are not next to each other?\\
For the second photograph, the members will stand in two rows, with 5 in the back row and 4 in the front row.
\item In how many different ways can the 9 members be divided into a group of 5 and a group of 4?
\item For a random division into a group of 5 and a group of 4, find the probability that Raman and Sanjay are in the same group as each other.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q5 [10]}}