| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Moderate -0.3 Part (a) is a straightforward application of the multiplication principle with combinations: C(4,1) × C(11,5). Part (b) requires complementary counting (total groups minus groups with both sisters), which is a standard technique but adds mild problem-solving beyond pure recall. Overall slightly easier than average A-level due to being a direct application of standard combinatorics methods with no novel insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^{11}C_5 \times\, ^4C_1\) | M1 | \(^{11}C_5 \times\, ^4C_1\) condone \(^{11}P_5 \times\, ^4P_1\) no \(+, -, \times\) or \(\div\) |
| \(1848\) | A1 | CAO as exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: [Neither selected \(=\)] \(^{13}C_6 [= 1716]\); [Only Jane selected \(=\)] \(^{13}C_5 [= 1287]\); [Only Kate selected \(=\)] \(^{13}C_5 [= 1287]\) | M1 | Either \(^{13}C_6\) seen alone or \(^{13}C_5\) seen alone or \(\times 2\) (condone \(^{13}P_n\), \(n = 5,6\)) |
| \([\text{Total} =] 1716 + 1287 + 1287\) | M1 | Three correct scenarios only added, accept unsimplified (values may be incorrect) |
| \(4290\) | A1 | |
| Method 2: \(^{15}C_6 - ^{13}C_4 [= 5005 - 715]\) | M1 | \(^{15}C_6 - k\), \(k\) a positive integer \(< 5005\), condone \(^{15}P_6\) |
| M1 | \(m - ^{13}C_4\), \(m\) integer \(> 715\), condone \(n - ^{13}P_4\), \(n > 17160\) | |
| \(4290\) | A1 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^{11}C_5 \times\, ^4C_1$ | M1 | $^{11}C_5 \times\, ^4C_1$ condone $^{11}P_5 \times\, ^4P_1$ no $+, -, \times$ or $\div$ |
| $1848$ | A1 | CAO as exact |
**Total: 2 marks**
---
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** [Neither selected $=$] $^{13}C_6 [= 1716]$; [Only Jane selected $=$] $^{13}C_5 [= 1287]$; [Only Kate selected $=$] $^{13}C_5 [= 1287]$ | M1 | Either $^{13}C_6$ seen alone or $^{13}C_5$ seen alone or $\times 2$ (condone $^{13}P_n$, $n = 5,6$) |
| $[\text{Total} =] 1716 + 1287 + 1287$ | M1 | Three correct scenarios only added, accept unsimplified (values may be incorrect) |
| $4290$ | A1 | |
| **Method 2:** $^{15}C_6 - ^{13}C_4 [= 5005 - 715]$ | M1 | $^{15}C_6 - k$, $k$ a positive integer $< 5005$, condone $^{15}P_6$ |
| | M1 | $m - ^{13}C_4$, $m$ integer $> 715$, condone $n - ^{13}P_4$, $n > 17160$ |
| $4290$ | A1 | |
**Total: 3 marks**
---2 A group of 6 people is to be chosen from 4 men and 11 women.
\begin{enumerate}[label=(\alph*)]
\item In how many different ways can a group of 6 be chosen if it must contain exactly 1 man?\\
Two of the 11 women are sisters Jane and Kate.
\item In how many different ways can a group of 6 be chosen if Jane and Kate cannot both be in the group?
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q2 [5]}}