| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with identical objects |
| Difficulty | Standard +0.3 This is a standard permutations with identical objects question requiring systematic case-by-case counting. Part (i) involves straightforward casework (cola/green tea/orange at ends), and part (ii) combines treating cola as a block with complementary counting for the green tea constraint. While multi-step, these are textbook techniques for this topic with no novel insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) ends cola, \(\frac{5!}{2!2!} = 30\) ends green tea, \(\frac{5!}{3!2!} = 10\) ends orange juice, \(\frac{5!}{3!2!} = 10\) total = 50 ways | M1 | Considering all three options |
| A1 | Any one option correct | |
| A1 | Correct answer | |
| OR \(P(\text{ends same}) = \frac{3}{7} \times \frac{2}{6} - \frac{2}{7} \times \frac{1}{6} + \frac{2}{7} \times \frac{1}{6} = \frac{5}{21}\) | M1 | OR Considering all three options |
| A1 | Correct fraction | |
| \(\frac{5}{21} \times \frac{7!}{3!2!2!} = 50 \text{ ways}\) | A1 | Correct answer |
| (ii) colas together, no restrictions, \(\frac{5!}{2!2!} = 30\) ways colas together and green tea together, \(\frac{4!}{2!} = 12\) ways \(30 - 12 = 18 \text{ ways}\) | M1 | Considering all colas together, or 5! seen |
| A1 | Correct answer | |
| M1 | Considering all colas tog and all green tea tog, or 4! seen | |
| A1 | Correct answer | |
| A1 | Correct final answer | |
| OR Attempt to list | M1A1 | OR₁ 10 or more, 12 or more correct |
| M1A1 | 14 or more, 16 or more correct | |
| A1 | 18 correct | |
| OR₂ \(3 \times \frac{4 \times 3}{2} = 18\) | M1 | OR₂ Considering all colas together, or 3! seen |
| A1 | 3 ways for colas and orange juice | |
| M1 | Considering green teas not together | |
| A1 | \(4 \times 3\) or \((4 \times 3)/2\) | |
| A1 | Correct final answer |
**(i)** ends cola, $\frac{5!}{2!2!} = 30$ ends green tea, $\frac{5!}{3!2!} = 10$ ends orange juice, $\frac{5!}{3!2!} = 10$ total = 50 ways | M1 | Considering all three options |
| A1 | Any one option correct |
| A1 | Correct answer |
OR $P(\text{ends same}) = \frac{3}{7} \times \frac{2}{6} - \frac{2}{7} \times \frac{1}{6} + \frac{2}{7} \times \frac{1}{6} = \frac{5}{21}$ | M1 | OR Considering all three options |
| A1 | Correct fraction |
$\frac{5}{21} \times \frac{7!}{3!2!2!} = 50 \text{ ways}$ | A1 | Correct answer |
**(ii)** colas together, no restrictions, $\frac{5!}{2!2!} = 30$ ways colas together and green tea together, $\frac{4!}{2!} = 12$ ways $30 - 12 = 18 \text{ ways}$ | M1 | Considering all colas together, or 5! seen |
| A1 | Correct answer |
| M1 | Considering all colas tog and all green tea tog, or 4! seen |
| A1 | Correct answer |
| A1 | Correct final answer |
OR Attempt to list | M1A1 | OR₁ 10 or more, 12 or more correct |
| M1A1 | 14 or more, 16 or more correct |
| A1 | 18 correct |
OR₂ $3 \times \frac{4 \times 3}{2} = 18$ | M1 | OR₂ Considering all colas together, or 3! seen |
| A1 | 3 ways for colas and orange juice |
| M1 | Considering green teas not together |
| A1 | $4 \times 3$ or $(4 \times 3)/2$ |
| A1 | Correct final answer |
---4 Three identical cans of cola, 2 identical cans of green tea and 2 identical cans of orange juice are arranged in a row. Calculate the number of arrangements if\\
(i) the first and last cans in the row are the same type of drink,\\
(ii) the 3 cans of cola are all next to each other and the 2 cans of green tea are not next to each other.
\hfill \mbox{\textit{CAIE S1 2010 Q4 [8]}}