CAIE S1 2019 November — Question 3 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2019
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCombinations & Selection
TypeMulti-stage selection problems
DifficultyStandard +0.3 This is a standard combinations problem with constraints requiring systematic case-by-case enumeration (part i) and a straightforward application of the division principle (part ii). While it requires careful organization and multiple calculations, the techniques are routine for A-level statistics students and involve no novel problem-solving insights.
Spec5.01b Selection/arrangement: probability problems

3 A sports team of 7 people is to be chosen from 6 attackers, 5 defenders and 4 midfielders. The team must include at least 3 attackers, at least 2 defenders and at least 1 midfielder.
  1. In how many different ways can the team of 7 people be chosen?
    The team of 7 that is chosen travels to a match in two cars. A group of 4 travel in one car and a group of 3 travel in the other car.
  2. In how many different ways can the team of 7 be divided into a group of 4 and a group of 3 ?

Question 3:
Part (i)
\(3A\ 2D\ 2M: {}^6C_3 \times {}^5C_2 \times {}^4C_2 (= 1200)\)
\(4A\ 2D\ 1M: {}^6C_4 \times {}^5C_2 \times {}^4C_1 (= 600)\)
AnswerMarks Guidance
\(3A\ 3D\ 1M: {}^6C_3 \times {}^5C_3 \times {}^4C_1 (= 800)\)M1 \({}^6C_x \times {}^5C_y \times {}^4C_z,\ x+y+z=7\)
A12 correct products, allow unsimplified
M1Summing their totals for 3 correct scenarios only
\(\text{Total} = 2600\)A1 Correct answer; SC1 \({}^6C_3 \times {}^5C_2 \times {}^4C_1 \times {}^9C_1 = 7200\)
Question 3(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(^7C_4 \times 1\)B1 \(^7C_3\) or \(^7C_4\) seen anywhere
\(35\)B1
2
## Question 3:

### Part (i)
$3A\ 2D\ 2M: {}^6C_3 \times {}^5C_2 \times {}^4C_2 (= 1200)$

$4A\ 2D\ 1M: {}^6C_4 \times {}^5C_2 \times {}^4C_1 (= 600)$

$3A\ 3D\ 1M: {}^6C_3 \times {}^5C_3 \times {}^4C_1 (= 800)$ | **M1** | ${}^6C_x \times {}^5C_y \times {}^4C_z,\ x+y+z=7$

| **A1** | 2 correct products, allow unsimplified

| **M1** | Summing their totals for 3 correct scenarios only

$\text{Total} = 2600$ | **A1** | Correct answer; **SC1** ${}^6C_3 \times {}^5C_2 \times {}^4C_1 \times {}^9C_1 = 7200$

## Question 3(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $^7C_4 \times 1$ | B1 | $^7C_3$ or $^7C_4$ seen anywhere |
| $35$ | B1 | |
| | **2** | |

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3 A sports team of 7 people is to be chosen from 6 attackers, 5 defenders and 4 midfielders. The team must include at least 3 attackers, at least 2 defenders and at least 1 midfielder.\\
\begin{enumerate}[label=(\roman*)]
\item In how many different ways can the team of 7 people be chosen?\\

The team of 7 that is chosen travels to a match in two cars. A group of 4 travel in one car and a group of 3 travel in the other car.
\item In how many different ways can the team of 7 be divided into a group of 4 and a group of 3 ?
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2019 Q3 [6]}}