CAIE S1 2019 June — Question 8 9 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2019
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCombinations & Selection
TypeMulti-stage selection problems
DifficultyModerate -0.3 This is a straightforward combinations question with standard techniques: (i) basic selection C(9,4), (ii) conditional selection with fixed items, (iii) treating buses as a block, (iv) arranging with restrictions. All parts use routine methods taught in S1 with no novel insight required, making it slightly easier than average but not trivial due to the multi-part nature.
Spec5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems

8 Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him.
  1. In how many different ways can Freddie choose 4 toys?
  2. How many of these choices will include both his favourite car and his favourite bus?
    Freddie arranges these 9 toys in a line.
  3. Find the number of possible arrangements if the buses are all next to each other.
  4. Find the number of possible arrangements if there is a car at each end of the line and no buses are next to each other.
    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 8(i):
AnswerMarks Guidance
AnswerMark Guidance
\(({}^9C_4 =)\ 126\)B1
Question 8(ii):
AnswerMarks Guidance
AnswerMark Guidance
\({}^7C_2\)B1 \({}^7C_x\) or \({}^3C_2\) (implied by correct answer) or \({}^7P_x\) or \({}^7P_y\), seen alone
\(= 21\)B1 Correct answer
Question 8(iii):
AnswerMarks Guidance
AnswerMark Guidance
\(\_\ C_1\ (B_1\ B_2\ B_3)\ C_2\ \_\ C_3\ \_\ C_4\ \_\ C_5\ \_\ C_6\)B1 3! or 6! seen alone or multiplied by \(k > 1\), need not be an integer
\(3! \times 6! \times 7\)B1 3! and 6! seen multiplied by \(k > 1\), integer, no division
\(= 30240\)B1 Exact value
Alternative: \(C_1\ (B_1\ B_2\ B_3)\ C_2\ C_3\ C_4\ C_5\ C_6\)
\(3! \times 7!\)B1 3! and 7! seen multiplied by \(k > \text{ or } = 1\), no division
\(= 30240\)B1 Exact value
Question 8(iv):
AnswerMarks Guidance
AnswerMark Guidance
\(C_1\ \_\ C_2\ \_\ C_3\ \_\ C_4\ \_\ C_5\ \_C_6\)B1 6! or 4! \(\times\) 6P2 seen alone or multiplied by \(k > 1\), no division (arrangements of cars)
\(6! \times 5\text{P}3\) or \(6! \times 5 \times 4 \times 3\) or \(6! \times 3! \times 10\)B1 Multiply by 5P3 oe i.e. putting Bs in between 4 of the Cs OR multiply by \(3! \times n\) where \(n = 7, 8, 9, 10\) (number of options)
\(= 43200\)B1 Correct answer
## Question 8(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $({}^9C_4 =)\ 126$ | B1 | |

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## Question 8(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| ${}^7C_2$ | B1 | ${}^7C_x$ or ${}^3C_2$ (implied by correct answer) or ${}^7P_x$ or ${}^7P_y$, seen alone |
| $= 21$ | B1 | Correct answer |

---

## Question 8(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\_\ C_1\ (B_1\ B_2\ B_3)\ C_2\ \_\ C_3\ \_\ C_4\ \_\ C_5\ \_\ C_6$ | B1 | 3! or 6! seen alone or multiplied by $k > 1$, need not be an integer |
| $3! \times 6! \times 7$ | B1 | 3! and 6! seen multiplied by $k > 1$, integer, no division |
| $= 30240$ | B1 | Exact value |
| **Alternative:** $C_1\ (B_1\ B_2\ B_3)\ C_2\ C_3\ C_4\ C_5\ C_6$ | | |
| $3! \times 7!$ | B1 | 3! and 7! seen multiplied by $k > \text{ or } = 1$, no division |
| $= 30240$ | B1 | Exact value |

---

## Question 8(iv):

| Answer | Mark | Guidance |
|--------|------|----------|
| $C_1\ \_\ C_2\ \_\ C_3\ \_\ C_4\ \_\ C_5\ \_C_6$ | B1 | 6! or 4! $\times$ 6P2 seen alone or multiplied by $k > 1$, no division (arrangements of cars) |
| $6! \times 5\text{P}3$ or $6! \times 5 \times 4 \times 3$ or $6! \times 3! \times 10$ | B1 | Multiply by 5P3 oe i.e. putting Bs in between 4 of the Cs OR multiply by $3! \times n$ where $n = 7, 8, 9, 10$ (number of options) |
| $= 43200$ | B1 | Correct answer |
8 Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him.\\
(i) In how many different ways can Freddie choose 4 toys?\\

(ii) How many of these choices will include both his favourite car and his favourite bus?\\

Freddie arranges these 9 toys in a line.\\
(iii) Find the number of possible arrangements if the buses are all next to each other.\\

(iv) Find the number of possible arrangements if there is a car at each end of the line and no buses are next to each other.\\

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

\hfill \mbox{\textit{CAIE S1 2019 Q8 [9]}}