| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Moderate -0.3 This is a standard permutations question with typical constraints (fixed positions, exclusions, conditional selections). Part (a) is straightforward arrangement with fixed ends, (b) uses complement counting, and (c) requires systematic case enumeration. All techniques are routine for S1 level with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part nature and need for careful counting of repeated letters. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{8!}{3!}\) | M1 | |
| \(6720\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total number = \(\frac{10!}{2!3!}\) (302400) | B1 | |
| With Es together = \(\frac{9!}{3!}\) (60480) | B1 | |
| Es not together = *their* (A) – *their* (B) | M1 | |
| 241920 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{8!}{3!} \times \frac{9 \times 8}{2}\) | ||
| \(8! \times k\) in numerator, \(k\) integer \(\geq 1\), denominator \(\geq 1\) | B1 | |
| \(3! \times m\) in denominator, \(m\) integer \(\geq 1\) | B1 | |
| *Their* \(\frac{8!}{3!}\) multiplied by \(^9C_2\) (OE) only (no additional terms) | M1 | |
| 241920 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Scenarios: EMMM \(^5C_0 = 1\); EMM_ \(^5C_1 = 5\); EM__ \(^5C_2 = 10\) | M1 | |
| Summing the number of ways for 2 or 3 correct scenarios | M1 | |
| Total \(= 16\) | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{8!}{3!}$ | M1 | |
| $6720$ | A1 | |
# Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Total number = $\frac{10!}{2!3!}$ (302400) | B1 | |
| With Es together = $\frac{9!}{3!}$ (60480) | B1 | |
| Es not together = *their* (A) – *their* (B) | M1 | |
| 241920 | A1 | |
**Alternative method for question 6(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{8!}{3!} \times \frac{9 \times 8}{2}$ | | |
| $8! \times k$ in numerator, $k$ integer $\geq 1$, denominator $\geq 1$ | B1 | |
| $3! \times m$ in denominator, $m$ integer $\geq 1$ | B1 | |
| *Their* $\frac{8!}{3!}$ multiplied by $^9C_2$ (OE) only (no additional terms) | M1 | |
| 241920 | A1 | |
---
# Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Scenarios: EMMM $^5C_0 = 1$; EMM_ $^5C_1 = 5$; EM__ $^5C_2 = 10$ | M1 | |
| Summing the number of ways for 2 or 3 correct scenarios | M1 | |
| Total $= 16$ | A1 | |
---6
\begin{enumerate}[label=(\alph*)]
\item Find the number of different ways in which the 10 letters of the word SUMMERTIME can be arranged so that there is an E at the beginning and an E at the end.
\item Find the number of different ways in which the 10 letters of the word SUMMERTIME can be arranged so that the Es are not together.
\item Four letters are selected from the 10 letters of the word SUMMERTIME. Find the number of different selections if the four letters include at least one M and exactly one E .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q6 [9]}}