| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | People arrangements in lines |
| Difficulty | Standard +0.3 This is a standard permutations question with three parts requiring familiar techniques: (i) treating families as blocks (4! × products of internal arrangements), (ii) arranging with restrictions (place children first, then adults in gaps), and (iii) dividing into groups with identical groups requiring division by factorials. All are textbook methods with straightforward application, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4! \times 3! \times 5! \times 2! \times 4! = 829440\) | B1 | 4!, 3!, 5!, 2 seen multiplied; 1 not in denominator |
| B1 | Mult by 4! | |
| B1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8! \times 9 \times 8 \times 7 \times 6 \times 5 \times 4\) | B1 | 8! seen multiplied; 1 |
| \(= 2438553600\ (2.44 \times 10^9)\) | B1 | Mult by \(_9P_6\) |
| B1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8C3 \times 5C3 \times 2C2\) | B1 | 8C3 seen mult |
| \(= 560\) | B1 | 5C3 seen mult |
| B1 [3] | Correct answer |
## Question 6:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4! \times 3! \times 5! \times 2! \times 4! = 829440$ | B1 | 4!, 3!, 5!, 2 seen multiplied; 1 not in denominator |
| | B1 | Mult by 4! |
| | B1 **[3]** | Correct answer |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $8! \times 9 \times 8 \times 7 \times 6 \times 5 \times 4$ | B1 | 8! seen multiplied; 1 |
| $= 2438553600\ (2.44 \times 10^9)$ | B1 | Mult by $_9P_6$ |
| | B1 **[3]** | Correct answer |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $8C3 \times 5C3 \times 2C2$ | B1 | 8C3 seen mult |
| $= 560$ | B1 | 5C3 seen mult |
| | B1 **[3]** | Correct answer |
---6 Four families go to a theme park together. Mr and Mrs Lin take their 2 children. Mr O'Connor takes his 2 children. Mr and Mrs Ahmed take their 3 children. Mrs Burton takes her son. The 14 people all have to go through a turnstile one at a time to enter the theme park.\\
(i) In how many different orders can the 14 people go through the turnstile if each family stays together?\\
(ii) In how many different orders can the 8 children and 6 adults go through the turnstile if no two adults go consecutively?
Once inside the theme park, the children go on the roller-coaster. Each roller-coaster car holds 3 people.\\
(iii) In how many different ways can the 8 children be divided into two groups of 3 and one group of 2 to go on the roller-coaster?
\hfill \mbox{\textit{CAIE S1 2013 Q6 [9]}}