| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Selection with family/relationship restrictions |
| Difficulty | Standard +0.3 This is a straightforward combinations problem with standard restrictions. Part (a) requires casework (4W+2M, 5W+1M, 6W+0M) using basic C(n,r) calculations. Part (b) uses complement counting (total minus both siblings present), a common technique. Both parts are routine applications of selection methods with no novel insight required, making this slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 6W 0M: \(^9C_6 = 84\) | M1 | Correct number of ways for either 5 or 4 women, accept unsimplified |
| 5W 1M: \(^9C_5 \times {^5C_1} = 126 \times 5 = 630\) | ||
| 4W 2M: \(^9C_4 \times {^5C_2} = 126 \times 10 = 1260\) | M1 | Summing the number of ways for 2 or 3 correct scenarios, no incorrect scenarios |
| Total \(= 1974\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total number of ways \(= {^{14}C_6}\) (3003) | M1 | \(^{14}C_6\) − a value |
| Number with sister and brother \(= {^{12}C_4}\) (495) | ||
| Number required \(= {^{14}C_6} -\ {^{12}C_4} = 3003 - 495\) | M1 | \(^{12}C_x\) or \(^nC_4\) seen on its own or subtracted from their total, \(x \leqslant 6\), \(n \leqslant 13\) |
| 2508 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Number of ways with neither \(= {^{12}C_6} = 924\) | M1 | \(^{12}C_6\) + a value |
| Number of ways with either brother or sister (not both) \(= {^{12}C_5} \times 2 = (792 \times 2) = 1584\) | M1 | \(^{12}C_x \times 2\) or \(^nC_5 \times 2\) seen on its own or added to their number of ways with neither, \(x \leqslant 5\), \(n \leqslant 12\) |
| Number required \(= 924 + 1584 = 2508\) | A1 |
## Question 3:
### Part 3(a):
6W 0M: $^9C_6 = 84$ | M1 | Correct number of ways for either 5 or 4 women, accept unsimplified
5W 1M: $^9C_5 \times {^5C_1} = 126 \times 5 = 630$ | |
4W 2M: $^9C_4 \times {^5C_2} = 126 \times 10 = 1260$ | M1 | Summing the number of ways for 2 or 3 correct scenarios, no incorrect scenarios
Total $= 1974$ | A1 |
### Part 3(b):
Total number of ways $= {^{14}C_6}$ (3003) | M1 | $^{14}C_6$ − a value
Number with sister and brother $= {^{12}C_4}$ (495) | |
Number required $= {^{14}C_6} -\ {^{12}C_4} = 3003 - 495$ | M1 | $^{12}C_x$ or $^nC_4$ seen on its own or subtracted from their total, $x \leqslant 6$, $n \leqslant 13$
2508 | A1 |
**Alternative method for 3(b):**
Number of ways with neither $= {^{12}C_6} = 924$ | M1 | $^{12}C_6$ + a value
Number of ways with either brother or sister (not both) $= {^{12}C_5} \times 2 = (792 \times 2) = 1584$ | M1 | $^{12}C_x \times 2$ or $^nC_5 \times 2$ seen on its own or added to their number of ways with neither, $x \leqslant 5$, $n \leqslant 12$
Number required $= 924 + 1584 = 2508$ | A1 |
---3 A committee of 6 people is to be chosen from 9 women and 5 men.
\begin{enumerate}[label=(\alph*)]
\item Find the number of ways in which the 6 people can be chosen if there must be more women than men on the committee.\\
The 9 women and 5 men include a sister and brother.
\item Find the number of ways in which the committee can be chosen if the sister and brother cannot both be on the committee.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q3 [6]}}