| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.8 This is a multi-part combinatorics problem requiring careful consideration of constraints in each part: distributing people among distinguishable cars with fixed drivers, arranging groups with ordering constraints, and selecting teams with family-based restrictions. While the individual techniques are standard A-level, the problem requires sustained careful reasoning across multiple scenarios and the potential for overcounting errors (especially in parts a and c) elevates it above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^{12}C_5 \times {}^7C_4\ [\times {}^3C_3]\) | M1 | \(^{12}C_r \times q\), \(r = 3, 4, 5\); \(q\) a positive integer \(> 1\), no \(+\) or \(-\) |
| M1 | \(^{12}C_s \times {}^{12-s}C_t\ [\times {}^{12-s-t}C_u]\); \(s = 3,4,5\); \(t = 3,4,5 \neq s\); \(u = 3,4,5 \neq s,t\) | |
| Alternative method: \(\dfrac{12!}{5! \times 3! \times 4!}\) | M1 | \(12! \div\) by a product of three factorials. |
| M1 | \(\dfrac{n!}{5! \times 3! \times 4!}\) | |
| [\(792 \times 35 =\)] \(27\,720\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4!\ \text{(Lizo)} \times 6!\ \text{(Kenny)} \times 2!\ \text{(Martin)} \times 2!\ \text{(Nantes)}\) | M1 | Product involving at least 3 of \(4!\), \(6!\), \(2!\), \(2!\) |
| \(\times\ 3!\) (orders of K, M and N) | M1 | \(w \times 3!\), \(w\) integer \(> 1\) |
| \(414\,720\) | A1 | WWW CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^7C_4\ \text{(adults)} \times {}^4C_1 \times {}^3C_1\) | M1 | \(^7C_4 \times b\), \(b\) integer \(> 1\), no \(+\) or \(-\) |
| \(420\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| K not L: \(^5C_3 \times {}^8C_3 = 560\) | M1 | \(^8C_3\) (or \(^8P_3\)) \(\times\ c\) for one of the products, or \(^5C_3\) (or \(^5P_3\)) \(\times\ c\), positive integer \(>1\) for first 2 products only. |
| L not K: \(^5C_3 \times {}^8C_3 = 560\) | ||
| L and K: \(^5C_2 \times {}^8C_3 = 560\) | M1 | Add 2 or 3 correct scenarios only values, no additional incorrect scenarios, no repeated scenarios. Accept unsimplified. |
| [Total \(=\)] \(1680\) | A1 | |
| Alternative: Total \(= {}^7C_4 \times {}^8C_3 = 1960\); Neither K nor L \(= {}^5C_4 \times {}^8C_3 = 280\) | M1, M1 | \(^8C_3 \times c\), \(c\) positive integer \(>1\). Subtracting the number of ways with neither from their total number of ways. |
| [Total \(=\)] \(1680\) | A1 | |
| Alternative: K: \(^6C_3 \times {}^8C_3 = 1120\); L: \(^6C_3 \times {}^8C_3 = 1120\); L and K: \(^5C_2 \times {}^8C_3 = 560\); \(1120 + 1120 - 560 = 1680\) | M1, M1 | \(^8C_3 \times c\), \(c\) positive integer \(>1\). Subtracting number of ways with both from sum of number of ways with K and number of ways with L. |
| [Total \(=\)] \(1680\) | A1 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $^{12}C_5 \times {}^7C_4\ [\times {}^3C_3]$ | M1 | $^{12}C_r \times q$, $r = 3, 4, 5$; $q$ a positive integer $> 1$, no $+$ or $-$ |
| | M1 | $^{12}C_s \times {}^{12-s}C_t\ [\times {}^{12-s-t}C_u]$; $s = 3,4,5$; $t = 3,4,5 \neq s$; $u = 3,4,5 \neq s,t$ |
| **Alternative method:** $\dfrac{12!}{5! \times 3! \times 4!}$ | M1 | $12! \div$ by a product of three factorials. |
| | M1 | $\dfrac{n!}{5! \times 3! \times 4!}$ |
| [$792 \times 35 =$] $27\,720$ | A1 | CAO |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4!\ \text{(Lizo)} \times 6!\ \text{(Kenny)} \times 2!\ \text{(Martin)} \times 2!\ \text{(Nantes)}$ | M1 | Product involving at least 3 of $4!$, $6!$, $2!$, $2!$ |
| $\times\ 3!$ (orders of K, M and N) | M1 | $w \times 3!$, $w$ integer $> 1$ |
| $414\,720$ | A1 | WWW CAO |
---
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $^7C_4\ \text{(adults)} \times {}^4C_1 \times {}^3C_1$ | M1 | $^7C_4 \times b$, $b$ integer $> 1$, no $+$ or $-$ |
| $420$ | A1 | |
---
## Question 7(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| K not L: $^5C_3 \times {}^8C_3 = 560$ | M1 | $^8C_3$ (or $^8P_3$) $\times\ c$ for one of the products, or $^5C_3$ (or $^5P_3$) $\times\ c$, positive integer $>1$ for first 2 products only. |
| L not K: $^5C_3 \times {}^8C_3 = 560$ | | |
| L and K: $^5C_2 \times {}^8C_3 = 560$ | M1 | Add 2 or 3 correct scenarios only values, no additional incorrect scenarios, no repeated scenarios. Accept unsimplified. |
| [Total $=$] $1680$ | A1 | |
| **Alternative:** Total $= {}^7C_4 \times {}^8C_3 = 1960$; Neither K nor L $= {}^5C_4 \times {}^8C_3 = 280$ | M1, M1 | $^8C_3 \times c$, $c$ positive integer $>1$. Subtracting the number of ways with neither from their total number of ways. |
| [Total $=$] $1680$ | A1 | |
| **Alternative:** K: $^6C_3 \times {}^8C_3 = 1120$; L: $^6C_3 \times {}^8C_3 = 1120$; L and K: $^5C_2 \times {}^8C_3 = 560$; $1120 + 1120 - 560 = 1680$ | M1, M1 | $^8C_3 \times c$, $c$ positive integer $>1$. Subtracting number of ways with both from sum of number of ways with K and number of ways with L. |
| [Total $=$] $1680$ | A1 | |7 A group of 15 friends visit an adventure park. The group consists of four families.
\begin{itemize}
\item Mr and Mrs Kenny and their four children
\item Mr and Mrs Lizo and their three children
\item Mrs Martin and her child
\item Mr and Mrs Nantes
\end{itemize}
The group travel to the park in three cars, one containing 6 people, one containing 5 people and one containing 4 people. The cars are driven by Mr Lizo, Mrs Martin and Mr Nantes respectively.
\begin{enumerate}[label=(\alph*)]
\item In how many different ways can the remaining 12 members of the group be divided between the three cars?\\
The group enter the park by walking through a gate one at a time.
\item In how many different orders can the 15 friends go through the gate if Mr Lizo goes first and each family stays together?\\
In the park, the group enter a competition which requires a team of 4 adults and 3 children.
\item In how many ways can the team be chosen from the group of 15 so that the 3 children are all from different families?
\item In how many ways can the team be chosen so that at least one of Mr Kenny or Mr Lizo is included?\\
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 2022 Q7 [11]}}