| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Committee with gender/category constraints |
| Difficulty | Standard +0.3 Part (a) is a straightforward application of the multiplication principle with two independent selections (C(6,3) × C(8,3)). Part (b) requires complementary counting or casework to handle the 'no more than two brothers' constraint, which is a standard technique but adds modest problem-solving demand beyond pure recall. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^6C_3 \times ^8C_3\) | M1 | \(^6C_3 \times b\) or \(c \times ^8C_3\) seen. \(b\), \(c\) integers \(\geq 1\) (1 may be implied) |
| \(1120\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 0 brothers: \([^3C_0] \times ^{11}C_6 = 462\); 1 brother: \(^3C_1 \times ^{11}C_5 = 1386\); 2 brothers: \(^3C_2 \times ^{11}C_4 = 990\) | B1 | \(^3C_x \times ^{11}C_{6-x}\), with \(x = 1\) or \(2\) seen |
| Add values of 3 correct scenarios | M1 | Add values of 3 correct scenarios (may be identified by appropriate calculations), no incorrect/repeated scenarios, condone use of permutations |
| \(2838\) | A1 | Only dependent on M mark. SC B1 for correct calculation or 2838 seen WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^{14}C_6 - ^{11}C_3\) | B1 | \(^{14}C_6 - d\), where \(d\) a positive integer |
| \(3003 - 165\) | M1 | \(e - ^{11}C_3\), where \(e\) is a positive integer \(> 165\) |
| \(= 2838\) | A1 | |
| 3 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^6C_3 \times ^8C_3$ | M1 | $^6C_3 \times b$ or $c \times ^8C_3$ seen. $b$, $c$ integers $\geq 1$ (1 may be implied) |
| $1120$ | A1 | |
| | **2** | |
## Question 2(b):
**Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| 0 brothers: $[^3C_0] \times ^{11}C_6 = 462$; 1 brother: $^3C_1 \times ^{11}C_5 = 1386$; 2 brothers: $^3C_2 \times ^{11}C_4 = 990$ | B1 | $^3C_x \times ^{11}C_{6-x}$, with $x = 1$ or $2$ seen |
| Add values of 3 correct scenarios | M1 | Add values of 3 correct scenarios (may be identified by appropriate calculations), no incorrect/repeated scenarios, condone use of permutations |
| $2838$ | A1 | Only dependent on M mark. **SC B1** for correct calculation or 2838 seen WWW |
**Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^{14}C_6 - ^{11}C_3$ | B1 | $^{14}C_6 - d$, where $d$ a positive integer |
| $3003 - 165$ | M1 | $e - ^{11}C_3$, where $e$ is a positive integer $> 165$ |
| $= 2838$ | A1 | |
| | **3** | |
---2
\begin{enumerate}[label=(\alph*)]
\item Find the number of ways in which a committee of 6 people can be chosen from 6 men and 8 women if it must include 3 men and 3 women.\\
A different committee of 6 people is to be chosen from 6 men and 8 women. Three of the 6 men are brothers.
\item Find the number of ways in which this committee can be chosen if there are no restrictions on the numbers of men and women, but it must include no more than two of the brothers.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q2 [5]}}