| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with identical objects |
| Difficulty | Moderate -0.3 This is a multi-part permutations question with standard techniques: (i) uses basic counting with odd number constraint, (ii) applies separation principle for non-adjacent arrangements, and (iii) involves arrangements with identical objects and a grouping constraint. All parts use well-established methods taught in S1, requiring careful application rather than novel insight, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \({}^{10}C_1 + {}^{10}C_3 + {}^{10}C_5 + {}^{10}C_7 + {}^{10}C_9 = 512\) | M1, A1, A1 | Summing some \({}^{10}C\) combinations with odd numbers, all different; At least 3 correct unsimplified expressions; Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(6! \times 7 \times 6 \times 5 = 151200\) | B1, M1, A1 | \(6!\) seen; multiplying by \({}^7P_3\) o.e.; correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{12!}{(4!) \times 7!} = 3960\) | B1, M1, A1 | \(12!\) Seen; dividing by \(4!7!\); correct answer |
**(i) Binomial coefficients:**
${}^{10}C_1 + {}^{10}C_3 + {}^{10}C_5 + {}^{10}C_7 + {}^{10}C_9 = 512$ | M1, A1, A1 | Summing some ${}^{10}C$ combinations with odd numbers, all different; At least 3 correct unsimplified expressions; Correct answer | **[3]**
**(ii) Permutations:**
$6! \times 7 \times 6 \times 5 = 151200$ | B1, M1, A1 | $6!$ seen; multiplying by ${}^7P_3$ o.e.; correct answer | **[3]**
**(iii) Combinations:**
$\frac{12!}{(4!) \times 7!} = 3960$ | B1, M1, A1 | $12!$ Seen; dividing by $4!7!$; correct answer | **[3]**
---6 (i) Find the number of different ways that a set of 10 different mugs can be shared between Lucy and Monica if each receives an odd number of mugs.\\
(ii) Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a different design. Find in how many ways these 9 mugs can be arranged in a row if the china mugs are all separated from each other.\\
(iii) Another set consists of 3 identical red mugs, 4 identical blue mugs and 7 identical yellow mugs. These 14 mugs are placed in a row. Find how many different arrangements of the colours are possible if the red mugs are kept together.
\hfill \mbox{\textit{CAIE S1 2010 Q6 [9]}}