CAIE S1 2024 June — Question 7 10 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2024
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPermutations & Arrangements
TypeArrangements with positional constraints
DifficultyStandard +0.3 This is a standard permutations question with repeated elements. Part (a) is direct application of the formula for arrangements with repetition. Part (b) requires complementary counting (total minus all-together cases), which is routine for this topic. Part (c) involves basic probability with combinations. All techniques are textbook exercises requiring no novel insight, making it slightly easier than average.
Spec5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems
7 The eight digits \(1,2,2,3,4,4,4,5\) are arranged in a line.
  1. How many different arrangements are there of these 8 digits?
  2. Find the number of different arrangements of the 8 digits in which there is a 2 at the beginning, a 2 at the end and the three 4 s are not all together.
    Three digits are selected at random from the eight digits \(1,2,2,3,4,4,4,5\).
  3. Find the probability that the three digits are all different.
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Question 7(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(\left[\frac{8!}{2!3!} =\right] 3360\)B1
Question 7(b):
AnswerMarks Guidance
AnswerMarks Guidance
Number of arrangements with 2s at the end \(-\) number of arrangements with 2s at the end and the 4s together: \(2\_\_\_\_\_\_2 - 2\_2\_(444)\_\_2\)
\(\frac{6!}{3!} - 4!\)M1 \(\frac{6!}{3!} \times r - s\), \(r = 1, 2\) and \(s\) a positive integer (including 0).
B1\(4!\) seen either alone or in \(t - 4!\), \(t\) an integer value \(> 24\).
M1\(\frac{6!}{3!} \times r - 4! \times u\), \(r = 1, 2\) and \(u = 1, 2\).
\(= 96\)A1
Question 7(c):
AnswerMarks Guidance
AnswerMarks Guidance
Scenarios: (2s=0, 4s=0, others=3): \(^3C_3 = 1\); (2s=0, 4s=1, others=2): \(^3C_1 \times {^3C_2} = 9\); (2s=1, 4s=0, others=2): \(^2C_1 \times {^3C_2} = 6\); (2s=1, 4s=1, others=1): \(^2C_1 \times {^3C_1} \times {^3C_1} = 18\)M1 One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4.
A1Two correct outcomes evaluated, accept unsimplified.
[Total 34 ways]M1 Four correct scenarios added.
Total number of selections \(= {^8C_3} = 56\)B1 Used as denominator of probability expression.
Probability \(= \frac{34}{{^8C_3}} = \frac{17}{28} \approx 0.607\)A1
Question 7(c):
Method 2 – Combinations of 3 numbers
AnswerMarks Guidance
ScenarioCalculation Value
1,2,3\(^1C_1 \times ^2C_1 \times ^1C_1\) 2
1,2,4\(^1C_1 \times ^2C_1 \times ^3C_1\) 6
1,2,5\(^1C_1 \times ^2C_1 \times ^1C_1\) 2
1,3,4\(^1C_1 \times ^1C_1 \times ^3C_1\) 3
1,3,5\(^1C_1 \times ^1C_1 \times ^1C_1\) 1
1,4,5\(^1C_1 \times ^3C_1 \times ^1C_1\) 3
2,3,4\(^2C_1 \times ^1C_1 \times ^3C_1\) 6
2,3,5\(^2C_1 \times ^1C_1 \times ^1C_1\) 2
2,4,5\(^2C_1 \times ^3C_1 \times ^1C_1\) 6
3,4,5\(^1C_1 \times ^3C_1 \times ^1C_1\) 3
Total 34 ways\(^8C_3 = 56\) (B1)
\(\text{Probability} = \dfrac{34}{^8C_3} = \dfrac{17}{28} \approx 0.607\)(A1)
Method 3 – Ordered probability
AnswerMarks Guidance
ScenarioCalculation Value
1,2,3\(\frac{1}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3!\) \(\frac{12}{336}\)
1,2,4\(\frac{1}{8}\times\frac{2}{7}\times\frac{3}{6}\times 3!\) \(\frac{36}{336}\)
1,2,5\(\frac{1}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3!\) \(\frac{12}{336}\)
1,3,4\(\frac{1}{8}\times\frac{1}{7}\times\frac{3}{6}\times 3!\) \(\frac{18}{336}\)
1,3,5\(\frac{1}{8}\times\frac{1}{7}\times\frac{1}{6}\times 3!\) \(\frac{6}{336}\)
1,4,5\(\frac{1}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3!\) \(\frac{18}{336}\)
2,3,4\(\frac{2}{8}\times\frac{1}{7}\times\frac{3}{6}\times 3!\) \(\frac{36}{336}\)
2,3,5\(\frac{2}{8}\times\frac{1}{7}\times\frac{1}{6}\times 3!\) \(\frac{12}{336}\)
2,4,5\(\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3!\) \(\frac{36}{336}\)
3,4,5\(\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3!\) \(\frac{18}{336}\)
\(\text{Probability} = \dfrac{17}{28} \approx 0.607\)(B1) 336 or \(8\times7\times6\) seen as denominator
Method 4 – Complementary (subtracting scenarios containing three 4s)
AnswerMarks Guidance
ScenarioCalculation Value
444\(\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}\) \(\frac{6}{336}\)
445\(\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3\) \(\frac{18}{336}\)
443\(\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3\) \(\frac{18}{336}\)
442\(\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3\) \(\frac{36}{336}\)
441\(\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3\) \(\frac{18}{336}\)
225\(\frac{2}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3\) \(\frac{6}{336}\)
224\(\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3\) \(\frac{18}{336}\)
223\(\frac{2}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3\) \(\frac{6}{336}\)
221\(\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3\) \(\frac{6}{336}\)
\(\text{Probability} = 1 - \dfrac{132}{336} = \dfrac{204}{336} \approx 0.607\)(B1) 336 or \(8\times7\times6\) seen as denominator
Total: 5 marks
## Question 7(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{8!}{2!3!} =\right] 3360$ | B1 | |

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## Question 7(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Number of arrangements with 2s at the end $-$ number of arrangements with 2s at the end and the 4s together: $2\_\_\_\_\_\_2 - 2\_2\_(444)\_\_2$ | | |
| $\frac{6!}{3!} - 4!$ | M1 | $\frac{6!}{3!} \times r - s$, $r = 1, 2$ and $s$ a positive integer (including 0). |
| | B1 | $4!$ seen either alone or in $t - 4!$, $t$ an integer value $> 24$. |
| | M1 | $\frac{6!}{3!} \times r - 4! \times u$, $r = 1, 2$ and $u = 1, 2$. |
| $= 96$ | A1 | |

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## Question 7(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Scenarios: (2s=0, 4s=0, others=3): $^3C_3 = 1$; (2s=0, 4s=1, others=2): $^3C_1 \times {^3C_2} = 9$; (2s=1, 4s=0, others=2): $^2C_1 \times {^3C_2} = 6$; (2s=1, 4s=1, others=1): $^2C_1 \times {^3C_1} \times {^3C_1} = 18$ | M1 | One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. |
| | A1 | Two correct outcomes evaluated, accept unsimplified. |
| [Total 34 ways] | M1 | Four correct scenarios added. |
| Total number of selections $= {^8C_3} = 56$ | B1 | Used as denominator of probability expression. |
| Probability $= \frac{34}{{^8C_3}} = \frac{17}{28} \approx 0.607$ | A1 | |

## Question 7(c):

**Method 2 – Combinations of 3 numbers**

| Scenario | Calculation | Value | Mark | Guidance |
|----------|-------------|-------|------|----------|
| 1,2,3 | $^1C_1 \times ^2C_1 \times ^1C_1$ | 2 | **(M1)** | One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4 |
| 1,2,4 | $^1C_1 \times ^2C_1 \times ^3C_1$ | 6 | **(A1)** | Five correct outcomes evaluated, accept unsimplified |
| 1,2,5 | $^1C_1 \times ^2C_1 \times ^1C_1$ | 2 | **(M1)** | Ten correct scenarios added |
| 1,3,4 | $^1C_1 \times ^1C_1 \times ^3C_1$ | 3 | | |
| 1,3,5 | $^1C_1 \times ^1C_1 \times ^1C_1$ | 1 | | |
| 1,4,5 | $^1C_1 \times ^3C_1 \times ^1C_1$ | 3 | | |
| 2,3,4 | $^2C_1 \times ^1C_1 \times ^3C_1$ | 6 | | |
| 2,3,5 | $^2C_1 \times ^1C_1 \times ^1C_1$ | 2 | | |
| 2,4,5 | $^2C_1 \times ^3C_1 \times ^1C_1$ | 6 | | |
| 3,4,5 | $^1C_1 \times ^3C_1 \times ^1C_1$ | 3 | | |

Total 34 ways | $^8C_3 = 56$ | **(B1)** | $^8C_3$ or 56 as denominator of probability expression |

$\text{Probability} = \dfrac{34}{^8C_3} = \dfrac{17}{28} \approx 0.607$ | **(A1)** | |

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**Method 3 – Ordered probability**

| Scenario | Calculation | Value | Mark | Guidance |
|----------|-------------|-------|------|----------|
| 1,2,3 | $\frac{1}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3!$ | $\frac{12}{336}$ | **(M1)** | One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4 |
| 1,2,4 | $\frac{1}{8}\times\frac{2}{7}\times\frac{3}{6}\times 3!$ | $\frac{36}{336}$ | **(A1)** | Five correct outcomes evaluated, accept unsimplified |
| 1,2,5 | $\frac{1}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3!$ | $\frac{12}{336}$ | **(M1)** | Ten correct scenarios added |
| 1,3,4 | $\frac{1}{8}\times\frac{1}{7}\times\frac{3}{6}\times 3!$ | $\frac{18}{336}$ | | |
| 1,3,5 | $\frac{1}{8}\times\frac{1}{7}\times\frac{1}{6}\times 3!$ | $\frac{6}{336}$ | | |
| 1,4,5 | $\frac{1}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3!$ | $\frac{18}{336}$ | | |
| 2,3,4 | $\frac{2}{8}\times\frac{1}{7}\times\frac{3}{6}\times 3!$ | $\frac{36}{336}$ | | |
| 2,3,5 | $\frac{2}{8}\times\frac{1}{7}\times\frac{1}{6}\times 3!$ | $\frac{12}{336}$ | | |
| 2,4,5 | $\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3!$ | $\frac{36}{336}$ | | |
| 3,4,5 | $\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3!$ | $\frac{18}{336}$ | | |

$\text{Probability} = \dfrac{17}{28} \approx 0.607$ | **(B1)** | 336 or $8\times7\times6$ seen as denominator | **(A1)** | |

---

**Method 4 – Complementary (subtracting scenarios containing three 4s)**

| Scenario | Calculation | Value | Mark | Guidance |
|----------|-------------|-------|------|----------|
| 444 | $\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}$ | $\frac{6}{336}$ | **(M1)** | 1-1 correct calculation, unsimplified for an identified scenario not containing three 4s |
| 445 | $\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3$ | $\frac{18}{336}$ | **(A1)** | Five correct probabilities evaluated, accept unsimplified |
| 443 | $\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3$ | $\frac{18}{336}$ | **(M1)** | Nine correct scenarios subtracted |
| 442 | $\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3$ | $\frac{36}{336}$ | | |
| 441 | $\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3$ | $\frac{18}{336}$ | | |
| 225 | $\frac{2}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3$ | $\frac{6}{336}$ | | |
| 224 | $\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3$ | $\frac{18}{336}$ | | |
| 223 | $\frac{2}{8}\times\frac{2}{7}\times\frac{1}{6}\times 3$ | $\frac{6}{336}$ | | |
| 221 | $\frac{2}{8}\times\frac{3}{7}\times\frac{1}{6}\times 3$ | $\frac{6}{336}$ | | |

$\text{Probability} = 1 - \dfrac{132}{336} = \dfrac{204}{336} \approx 0.607$ | **(B1)** | 336 or $8\times7\times6$ seen as denominator | **(A1)** | |

**Total: 5 marks**
7 The eight digits $1,2,2,3,4,4,4,5$ are arranged in a line.
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements are there of these 8 digits?
\item Find the number of different arrangements of the 8 digits in which there is a 2 at the beginning, a 2 at the end and the three 4 s are not all together.\\

Three digits are selected at random from the eight digits $1,2,2,3,4,4,4,5$.
\item Find the probability that the three digits are all different.\\

If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2024 Q7 [10]}}
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