| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| 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 testing basic counting principles (treating letters as a block, complementary counting, and selections with restrictions). Part (i) is routine block arrangement, part (ii) uses standard complement method, and part (iii) is straightforward selection with constraints. While it requires careful handling of repeated letters (4 Es, 2 Ss) and multiple parts, these are textbook techniques for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{9!}{2!} = 181440\) | B1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total no. of ways \(= \frac{12!}{2!4!} = 9\,979\,200\) (A) | B1 | Accept unevaluated |
| With Ss together \(= \frac{11!}{4!} = 1\,663\,200\) (B) | B1 | Accept unevaluated |
| With Ss not together \(= (B) - (A)\) | M1 | Correct or \(\frac{12!}{m} - \frac{8!}{n}\), \(m,n\) integers \(>1\), or *their* identified total − *their* identified Ss together |
| \(8\,316\,000\) | A1 | Exact value |
| Alternative: \(\_T\_E\_E\_P\_L\_E\_C\_H\_A\_E\_\) | B1 | \(10! \times k\) in numerator, \(k\) integer \(\geqslant 1\) |
| \(\frac{10!}{4!} \times \frac{11 \times 10}{2!}\) | B1 | \(4! \times k\) in numerator, \(k\) integer \(\geqslant 1\) |
| \(\frac{\text{their } 10!}{\text{their } 4!} \times {}^{11}C_2\) or \({}^{11}P_2\) | M1 | OE |
| \(8\,316\,000\) | A1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(SEEE:1\) | M1 | \({}^6C_3\) seen alone or times \(K>1\) |
| \(SEE\_: {}^6C_1 = 6\); \(SE\_\_: {}^6C_2 = 15\); \(S\_\_\_: {}^6C_3 = 20\) | B1 | \({}^6C_3\) or \({}^6C_2\) or \({}^6C_1\) alone |
| Add 3 or 4 correct scenarios | M1 | No extras |
| Total \(= 42\) | A1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{9!}{2!} = 181440$ | B1 | Exact value |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total no. of ways $= \frac{12!}{2!4!} = 9\,979\,200$ (A) | B1 | Accept unevaluated |
| With Ss together $= \frac{11!}{4!} = 1\,663\,200$ (B) | B1 | Accept unevaluated |
| With Ss not together $= (B) - (A)$ | M1 | Correct or $\frac{12!}{m} - \frac{8!}{n}$, $m,n$ integers $>1$, or *their* identified total − *their* identified Ss together |
| $8\,316\,000$ | A1 | Exact value |
| **Alternative:** $\_T\_E\_E\_P\_L\_E\_C\_H\_A\_E\_$ | B1 | $10! \times k$ in numerator, $k$ integer $\geqslant 1$ |
| $\frac{10!}{4!} \times \frac{11 \times 10}{2!}$ | B1 | $4! \times k$ in numerator, $k$ integer $\geqslant 1$ |
| $\frac{\text{their } 10!}{\text{their } 4!} \times {}^{11}C_2$ or ${}^{11}P_2$ | M1 | OE |
| $8\,316\,000$ | A1 | Exact value |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $SEEE:1$ | M1 | ${}^6C_3$ seen alone or times $K>1$ |
| $SEE\_: {}^6C_1 = 6$; $SE\_\_: {}^6C_2 = 15$; $S\_\_\_: {}^6C_3 = 20$ | B1 | ${}^6C_3$ or ${}^6C_2$ or ${}^6C_1$ alone |
| Add 3 or 4 correct scenarios | M1 | No extras |
| Total $= 42$ | A1 | |
---6 (i) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that all four Es are together.\\
(ii) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that the Ss are not next to each other.\\
Four letters are selected from the 12 letters of the word STEEPLECHASE.\\
(iii) Find the number of different selections if the four letters include exactly one $S$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q6 [9]}}