| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with adjacency requirements |
| Difficulty | Moderate -0.3 This is a standard permutations question with repeated letters (SHOPKEEPER has 3 Es, 2 Ps) covering textbook techniques: treating items as a block (part a), complementary counting (part b), conditional probability with arrangements (part c), and combinations with restrictions (part d). While it requires careful bookkeeping across multiple parts, each individual technique is routine for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{8!}{2!}\) | M1 | \(\frac{8!}{k} \equiv \frac{7!\times8}{k}\), where \(k \in \mathbb{N}\); \(\frac{a!}{2(!)}\), where \(a \in \mathbb{N}\) |
| \(20160\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total number of ways: \(\frac{10!}{2!3!} (= 302400)\) (A) | B1 | Accept unsimplified |
| With Ps together: \(\frac{9!}{3!} (= 60480)\) (B) | B1 | Accept unsimplified |
| With Ps not together: \(302400 - 60480\) | M1 | \(\frac{10!}{m} - \frac{9!}{n}\), \(m,n\) integers or \((A)-(B)\) if clearly identified |
| \(241920\) | A1 | |
| Alternative method: | ||
| \(\frac{8!}{3!}\) | B1 | \(k \times 8!\) in numerator, \(k\) a positive integer, no \(\pm\) |
| \(\times\frac{9\times8}{2}\) | B1 | \(m \times 3!\) in denominator, \(m\) a positive integer, no \(\pm\) |
| M1 | *Their* \(\frac{8!}{3!}\) multiplied by \(^9C_2\) or \(^9P_2\), no additional terms | |
| \(241920\) | A1 | Exact value, WWW |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Probability} = \frac{\text{Number of ways Es at beginning and end}}{\text{Total number of ways}}\) | M1 | \(\dfrac{\left(\frac{8!}{k!}\right)}{\frac{10!}{k!l!}}\), \(1 \leqslant k,l \in \mathbb{N} \leqslant 3\), FT denominator from 7(b) or correct |
| \(\text{Probability} = \dfrac{\frac{8!}{2!}}{\frac{10!}{2\times3!}} = \frac{20160}{302400}\) | ||
| \(\frac{1}{15},\ 0.0667\) | A1 | |
| Alternative method: | ||
| \(\text{Probability} = \frac{3}{10}\times\frac{2}{9}\) | M1 | \(\frac{a}{10}\times\frac{a-1}{9},\ a = 3, 2\) |
| \(\frac{1}{15},\ 0.0667\) | A1 | |
| Alternative method: | ||
| \(\text{Probability} = \frac{1}{10}\times\frac{1}{9}\times3!\) | M1 | \(\frac{1}{10}\times\frac{1}{9}\times m!,\ m = 3,2\) |
| \(\frac{1}{15},\ 0.0667\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Scenarios: PEEE: \(^5C_0 = 1\); PEE\_: \(^5C_1 = 5\); PE\_\_: \(^5C_2 = 10\); P\_\_\_: \(^5C_3 = 10\) | M1 | \(^5C_x\) seen alone, \(1 \leqslant x \leqslant 4\) |
| M1 | Summing the number of ways for 3 or 4 correct scenarios (can be unsimplified), no incorrect scenarios | |
| Total \(= 26\) | A1 | |
| 3 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{8!}{2!}$ | M1 | $\frac{8!}{k} \equiv \frac{7!\times8}{k}$, where $k \in \mathbb{N}$; $\frac{a!}{2(!)}$, where $a \in \mathbb{N}$ |
| $20160$ | A1 | |
| | **2** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total number of ways: $\frac{10!}{2!3!} (= 302400)$ (A) | B1 | Accept unsimplified |
| With Ps together: $\frac{9!}{3!} (= 60480)$ (B) | B1 | Accept unsimplified |
| With Ps not together: $302400 - 60480$ | M1 | $\frac{10!}{m} - \frac{9!}{n}$, $m,n$ integers or $(A)-(B)$ if clearly identified |
| $241920$ | A1 | |
| **Alternative method:** | | |
| $\frac{8!}{3!}$ | B1 | $k \times 8!$ in numerator, $k$ a positive integer, no $\pm$ |
| $\times\frac{9\times8}{2}$ | B1 | $m \times 3!$ in denominator, $m$ a positive integer, no $\pm$ |
| | M1 | *Their* $\frac{8!}{3!}$ multiplied by $^9C_2$ or $^9P_2$, no additional terms |
| $241920$ | A1 | Exact value, WWW |
| | **4** | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Probability} = \frac{\text{Number of ways Es at beginning and end}}{\text{Total number of ways}}$ | M1 | $\dfrac{\left(\frac{8!}{k!}\right)}{\frac{10!}{k!l!}}$, $1 \leqslant k,l \in \mathbb{N} \leqslant 3$, FT denominator from **7(b)** or correct |
| $\text{Probability} = \dfrac{\frac{8!}{2!}}{\frac{10!}{2\times3!}} = \frac{20160}{302400}$ | | |
| $\frac{1}{15},\ 0.0667$ | A1 | |
| **Alternative method:** | | |
| $\text{Probability} = \frac{3}{10}\times\frac{2}{9}$ | M1 | $\frac{a}{10}\times\frac{a-1}{9},\ a = 3, 2$ |
| $\frac{1}{15},\ 0.0667$ | A1 | |
| **Alternative method:** | | |
| $\text{Probability} = \frac{1}{10}\times\frac{1}{9}\times3!$ | M1 | $\frac{1}{10}\times\frac{1}{9}\times m!,\ m = 3,2$ |
| $\frac{1}{15},\ 0.0667$ | A1 | |
| | **2** | |
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Scenarios: PEEE: $^5C_0 = 1$; PEE\_: $^5C_1 = 5$; PE\_\_: $^5C_2 = 10$; P\_\_\_: $^5C_3 = 10$ | M1 | $^5C_x$ seen alone, $1 \leqslant x \leqslant 4$ |
| | M1 | Summing the number of ways for 3 or 4 correct scenarios (can be unsimplified), no incorrect scenarios |
| Total $= 26$ | A1 | |
| | **3** | |7
\begin{enumerate}[label=(\alph*)]
\item Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that all 3 Es are together.
\item Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that the Ps are not next to each other.
\item Find the probability that a randomly chosen arrangement of the 10 letters of the word SHOPKEEPER has an E at the beginning and an E at the end.\\
Four letters are selected from the 10 letters of the word SHOPKEEPER.
\item Find the number of different selections if the four letters include exactly one P .\\
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 [11]}}