| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with identical objects |
| Difficulty | Moderate -0.3 This is a straightforward permutations question requiring standard techniques: (i) basic arrangement formula P(18,5), (ii) complement probability with consecutive arrangements, (iii) combinations with 'at least' condition. All parts use routine methods with no novel insight required, making it slightly easier than average but still requiring careful execution across multiple parts. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^{18}P_5\) | M1 | \(^{18}P_x\) or \(^yP_5\) OE seen, \(0 < x < 18\) and \(5 < y < 18\), can be mult by \(k \geq 1\) |
| \(= 1\,028\,160\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. \(*(\text{CCCCC})\) in \(5! \times 14\) ways \(= 1680\) | (B1) | \(5!\) OE mult by \(k \geq 1\), considering arrangements of cars next to each other |
| Multiply by 14 | B1 | Mult by 14 OE, (or 14 on its own) considering positions within the line |
| \(P(\text{next to each other}) = 1680/1\,028\,160\) | M1 | Dividing by (i) for probability |
| \(P(\text{not next to each other}) = 1 - 1680/1\,028\,160\) | M1 | Subtracting prob from 1 (or their \(5! \times 14\) from (i)) |
| \(= 0.998\left(\dfrac{611}{612}\right)\) OE | A1) | |
| *OR1:* \(\dfrac{5! \times 14!}{18!} = 0.001634\) | (B1) | \(5!\) OE mult by \(k \geq 1\) (on its own or in numerator of fraction) |
| Multiply by \(14!\) | B1 | Multiply by \(14!\), (or \(14!\) on its own) considering all ways of arranging spaces with 5 cars together |
| Dividing by \(18!\) | M1 | Dividing by \(18!\), total number of ways of arranging spaces |
| \(1 - 0.001634\) | M1 | Subtracting prob from 1 (or \(5! \times 14!\) from \(18!\)) |
| \(= 0.998(366)\) | A1) | |
| *OR2:* Sum of six scenarios \(= 1\,026\,480\) | (M1) | Listing the six correct scenarios (only): 4 together; 3 together and 2 separate; 3 together and 2 together; two sets of 2 together and 1 separate; 2 together and 3 separate; 5 separate. |
| Total \(= 1\,026\,480\) | A1 | Total of \(1\,026\,480\) |
| Dividing \(1\,026\,480\) by their 6(i) | M1 | Dividing their \(1\,026\,480\) by their 6(i) |
| \(1\,026\,480 \div 1\,028\,160 = 0.998(366)\) | A1) | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5C1 \times 4C1 \times 3C1\) | B1 | \(5C1 \times 4C1 \times 3C1\) or better seen i.e. no. of ways with 3 different colours |
| Any of \(^5C_2\) or \(^4C_2\) or \(^3C_2\) seen multiplied by \(k > 1\) | M1 | Any of \(^5C_2\) or \(^4C_2\) or \(^3C_2\) seen multiplied by \(k > 1\) (can be implied) |
| 2 correct unsimplified 'no. of ways' other than \(5C1 \times 4C1 \times 3C1\): \(^4C_2 \times 3 = 18\), \(5 \times {^3C_2} = 15\), \({^5C_2} \times 3 = 30\), \(5 \times {^4C_2} = 30\), \({^5C_2} \times 4 = 40\) | A1 | 2 correct unsimplified 'no. of ways' other than \(5C1 \times 4C1 \times 3C1\) |
| Summing no more than 7 scenario totals containing at least 6 correct scenarios | M1 | Summing no more than 7 scenario totals containing at least 6 correct scenarios |
| Total \(= 205\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^{12}C_3 -\) | M1 | Seeing \(^{12}C_3 -\), considering all selections of 3 cars |
| \(- {^5C_3}\) | M1 | Subt \(^5C_3\) OE, removing only red selections |
| \(- {^4C_3}\) | M1 | Subt \(^4C_3\) OE, removing only white selections |
| \(- {^3C_3}\) | M1 | Subt \(^3C_3\) OE, removing only black selections |
| \(= 205\) | A1 | Correct answer |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $^{18}P_5$ | M1 | $^{18}P_x$ or $^yP_5$ OE seen, $0 < x < 18$ and $5 < y < 18$, can be mult by $k \geq 1$ |
| $= 1\,028\,160$ | A1 | |
| | **2** | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. $***(\text{CCCCC})**********$ in $5! \times 14$ ways $= 1680$ | (B1) | $5!$ OE mult by $k \geq 1$, considering arrangements of cars next to each other |
| Multiply by 14 | B1 | Mult by 14 OE, (or 14 on its own) considering positions within the line |
| $P(\text{next to each other}) = 1680/1\,028\,160$ | M1 | Dividing by **(i)** for probability |
| $P(\text{not next to each other}) = 1 - 1680/1\,028\,160$ | M1 | Subtracting prob from 1 (or their $5! \times 14$ from **(i)**) |
| $= 0.998\left(\dfrac{611}{612}\right)$ OE | A1) | |
| *OR1:* $\dfrac{5! \times 14!}{18!} = 0.001634$ | (B1) | $5!$ OE mult by $k \geq 1$ (on its own or in numerator of fraction) |
| Multiply by $14!$ | B1 | Multiply by $14!$, (or $14!$ on its own) considering all ways of arranging spaces with 5 cars together |
| Dividing by $18!$ | M1 | Dividing by $18!$, total number of ways of arranging spaces |
| $1 - 0.001634$ | M1 | Subtracting prob from 1 (or $5! \times 14!$ from $18!$) |
| $= 0.998(366)$ | A1) | |
| *OR2:* Sum of six scenarios $= 1\,026\,480$ | (M1) | Listing the six correct scenarios (only): 4 together; 3 together and 2 separate; 3 together and 2 together; two sets of 2 together and 1 separate; 2 together and 3 separate; 5 separate. |
| Total $= 1\,026\,480$ | A1 | Total of $1\,026\,480$ |
| Dividing $1\,026\,480$ by their **6(i)** | M1 | Dividing their $1\,026\,480$ by their **6(i)** |
| $1\,026\,480 \div 1\,028\,160 = 0.998(366)$ | A1) | |
| | **5** | |
## Question 6(iii):
**Method 1 (Scenarios):**
R(5) W(4) B(3) scenarios:
| Answer | Mark | Guidance |
|--------|------|----------|
| $5C1 \times 4C1 \times 3C1$ | B1 | $5C1 \times 4C1 \times 3C1$ or better seen i.e. no. of ways with 3 different colours |
| Any of $^5C_2$ or $^4C_2$ or $^3C_2$ seen multiplied by $k > 1$ | M1 | Any of $^5C_2$ or $^4C_2$ or $^3C_2$ seen multiplied by $k > 1$ (can be implied) |
| 2 correct unsimplified 'no. of ways' other than $5C1 \times 4C1 \times 3C1$: $^4C_2 \times 3 = 18$, $5 \times {^3C_2} = 15$, ${^5C_2} \times 3 = 30$, $5 \times {^4C_2} = 30$, ${^5C_2} \times 4 = 40$ | A1 | 2 correct unsimplified 'no. of ways' other than $5C1 \times 4C1 \times 3C1$ |
| Summing no more than 7 scenario totals containing at least 6 correct scenarios | M1 | Summing no more than 7 scenario totals containing at least 6 correct scenarios |
| Total $= 205$ | A1 | |
**Method 2 (Complementary):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $^{12}C_3 -$ | M1 | Seeing $^{12}C_3 -$, considering all selections of 3 cars |
| $- {^5C_3}$ | M1 | Subt $^5C_3$ OE, removing only red selections |
| $- {^4C_3}$ | M1 | Subt $^4C_3$ OE, removing only white selections |
| $- {^3C_3}$ | M1 | Subt $^3C_3$ OE, removing only black selections |
| $= 205$ | A1 | Correct answer |
**Total: 5 marks**
---6 A car park has spaces for 18 cars, arranged in a line. On one day there are 5 cars, of different makes, parked in randomly chosen positions and 13 empty spaces.\\
(i) Find the number of possible arrangements of the 5 cars in the car park.\\
(ii) Find the probability that the 5 cars are not all next to each other.\\
On another day, 12 cars of different makes are parked in the car park. 5 of these cars are red, 4 are white and 3 are black. Elizabeth selects 3 of these cars.\\[0pt]
(iii) Find the number of selections Elizabeth can make that include cars of at least 2 different colours. [5]\\
\hfill \mbox{\textit{CAIE S1 2017 Q6 [12]}}