| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Moderate -0.8 Part (i) is a standard permutations with repetition problem (9!/(3!2!2!)). Part (ii) adds basic positional constraints but remains straightforward—fix D at start, then count arrangements for two cases (R or O at end). Both parts are routine textbook exercises requiring only direct application of formulas with minimal problem-solving. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{9!}{2!3!} = 30240\) | B1 | \(9!\) divided by at least one of \(2!\) or \(3!\) |
| B1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| \(D\text{-------}O: \frac{7!}{3!} = 840\) | B1 | \(7!\) seen alone or as numerator in a term, can be multiplied not \(+\) or \(-\) |
| B1 | One term correct, unsimplified | |
| \(\text{Total} = 2100\) | B1 | Final answer |
## Question 2:
### Part (i)
$\frac{9!}{2!3!} = 30240$ | **B1** | $9!$ divided by at least one of $2!$ or $3!$
| **B1** | Exact value
---
### Part (ii)
$D\text{-------}R: \frac{7!}{2!2!} = 1260$
$D\text{-------}O: \frac{7!}{3!} = 840$ | **B1** | $7!$ seen alone or as numerator in a term, can be multiplied not $+$ or $-$
| **B1** | One term correct, unsimplified
$\text{Total} = 2100$ | **B1** | Final answer
---2 (i) How many different arrangements are there of the 9 letters in the word CORRIDORS?\\
(ii) How many different arrangements are there of the 9 letters in the word CORRIDORS in which the first letter is D and the last letter is R or O ?\\
\hfill \mbox{\textit{CAIE S1 2019 Q2 [5]}}