| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| 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 repeated letters and basic constraints. Part (a) is routine factorial division, (b) fixes positions reducing to simpler counting, (c) uses complement counting with adjacent letters, and (d) applies systematic case-by-case selection. All techniques are textbook exercises requiring no novel insight, making it slightly easier than average A-level difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{9!}{2!2!} = 90\,720\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{6!}{2!}\) | M1 | |
| \(360\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 2 Es together \(= \frac{8!}{2!} (= 20160)\) | M1 | |
| Es not together \(= 90720 - 20160 = 70560\) | M1 | |
| Probability \(= \frac{70560}{90720}\) | M1 | |
| \(\frac{7}{9}\) or \(0.778\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{7!}{2!} \times \frac{8 \times 7}{2} = 70560\) | ||
| \(7! \times k\) in numerator, \(k\) integer \(\geqslant 1\), denominator \(\geqslant 1\) | M1 | |
| Multiplying by \(^8C_2\) OE | M1 | |
| Probability \(= \frac{70560}{90720}\) | M1 | |
| \(\frac{7}{9}\) or \(0.778\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Scenarios: \(EL\_{\_\_}\ ^5C_3 = 10\); \(EEL\_{\_}\ ^5C_2 = 10\); \(E\_{\_\_\_}\ ^5C_4 = 5\); \(EE\_{\_\_}\ ^5C_3 = 10\) | M1 | |
| Summing the number of ways for 3 or 4 correct scenarios | M1 | |
| Total \(= 35\) | A1 |
## Question 7:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{9!}{2!2!} = 90\,720$ | B1 | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{6!}{2!}$ | M1 | |
| $360$ | A1 | |
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| 2 Es together $= \frac{8!}{2!} (= 20160)$ | M1 | |
| Es not together $= 90720 - 20160 = 70560$ | M1 | |
| Probability $= \frac{70560}{90720}$ | M1 | |
| $\frac{7}{9}$ or $0.778$ | A1 | |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{7!}{2!} \times \frac{8 \times 7}{2} = 70560$ | | |
| $7! \times k$ in numerator, $k$ integer $\geqslant 1$, denominator $\geqslant 1$ | M1 | |
| Multiplying by $^8C_2$ OE | M1 | |
| Probability $= \frac{70560}{90720}$ | M1 | |
| $\frac{7}{9}$ or $0.778$ | A1 | |
---
## Question 7(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Scenarios: $EL\_{\_\_}\ ^5C_3 = 10$; $EEL\_{\_}\ ^5C_2 = 10$; $E\_{\_\_\_}\ ^5C_4 = 5$; $EE\_{\_\_}\ ^5C_3 = 10$ | M1 | |
| Summing the number of ways for 3 or 4 correct scenarios | M1 | |
| Total $= 35$ | A1 | |7
\begin{enumerate}[label=(\alph*)]
\item Find the number of different possible arrangements of the 9 letters in the word CELESTIAL.
\item Find the number of different arrangements of the 9 letters in the word CELESTIAL in which the first letter is C, the fifth letter is T and the last letter is E.
\item Find the probability that a randomly chosen arrangement of the 9 letters in the word CELESTIAL does not have the two Es together.\\
5 letters are selected at random from the 9 letters in the word CELESTIAL.
\item Find the number of different selections if the 5 letters include at least one E and at most one L .\\
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 2020 Q7 [10]}}