| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with grouped categories |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question requiring only standard formulas. Part (i) uses the concept of treating groups as single units (3! × 3! × 4! × 8!), part (ii) is direct application of C(3,2) × C(4,2) × C(8,2), and part (iii) requires simple case enumeration. All techniques are routine S1 material with no novel problem-solving or insight required, making it easier than average A-level questions. |
| Spec | 5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3! \times 4! \times 8! \times 3!\) | M1 | Multiplying 3 factorials together |
| M1 | Multiplying by \(3!\) | |
| \(= 34\,836\,480\ (34\,800\,000)\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \({}^3C_2 \times {}^4C_2 \times {}^8C_2\) | M1 | Multiplying (only) 3 combinations together |
| \(= 504\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Fr=3, Fa=1, H=2: \({}^8C_3 \times {}^3C_1 \times {}^4C_2 = 1008\) | M1 | Multiplying 3 combinations, only |
| Fr=3, Fa=2, H=1: \({}^8C_3 \times {}^3C_2 \times {}^4C_1 = 672\) | M1 | Summing 3 options |
| Fr=4, Fa=1, H=1: \({}^8C_4 \times {}^3C_1 \times {}^4C_1 = 840\) | A1 | 3 correct combination answers |
| total ways \(= 2520\) | A1 [4] | Correct answer |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3! \times 4! \times 8! \times 3!$ | M1 | Multiplying 3 factorials together |
| | M1 | Multiplying by $3!$ |
| $= 34\,836\,480\ (34\,800\,000)$ | A1 [3] | Correct answer |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| ${}^3C_2 \times {}^4C_2 \times {}^8C_2$ | M1 | Multiplying (only) 3 combinations together |
| $= 504$ | A1 [2] | Correct answer |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Fr=3, Fa=1, H=2: ${}^8C_3 \times {}^3C_1 \times {}^4C_2 = 1008$ | M1 | Multiplying 3 combinations, only |
| Fr=3, Fa=2, H=1: ${}^8C_3 \times {}^3C_2 \times {}^4C_1 = 672$ | M1 | Summing 3 options |
| Fr=4, Fa=1, H=1: ${}^8C_4 \times {}^3C_1 \times {}^4C_1 = 840$ | A1 | 3 correct combination answers |
| total ways $= 2520$ | A1 [4] | Correct answer |
---4 Mary saves her digital images on her computer in three separate folders named 'Family', 'Holiday' and 'Friends'. Her family folder contains 3 images, her holiday folder contains 4 images and her friends folder contains 8 images. All the images are different.\\
(i) Find in how many ways she can arrange these 15 images in a row across her computer screen if she keeps the images from each folder together.\\
(ii) Find the number of different ways in which Mary can choose 6 of these images if there are 2 from each folder.\\
(iii) Find the number of different ways in which Mary can choose 6 of these images if there are at least 3 images from the friends folder and at least 1 image from each of the other two folders.
\hfill \mbox{\textit{CAIE S1 2011 Q4 [9]}}