| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.3 This is a straightforward multi-stage selection problem requiring systematic case enumeration in part (i) and a standard 'grouping' permutation in part (ii). Both parts use routine combinatorial techniques (combinations with constraints, treating grouped items as single units) that are well-practiced at this level, with no novel insight required beyond careful organization of cases. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Scenarios: \(4V + 1C + 1DB\): \(^{11}C_4 \times ^5C_1 \times ^4C_1\) | M1 | \(^{11}C_a \times ^5C_b \times ^4C_c\), \(a+b+c=6\) |
| \(4V + 2C\): \(^{11}C_4 \times ^5C_2\); \(5V + 1C\): \(^{11}C_5 \times ^5C_1\) | B1 | 2 correct unsimplified options |
| \(6600 + 3300 + 2310\) | M1 | Add 2 or 3 correct scenarios only |
| \(= 12210\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4! \times 3!\) | M1 | \(k\) multiplied by \(3!\) or \(4!\), \(k\) an integer \(\geq 1\) |
| A1 | Correct unsimplified expression | |
| \(= 144\) | A1 | Correct answer |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Scenarios: $4V + 1C + 1DB$: $^{11}C_4 \times ^5C_1 \times ^4C_1$ | M1 | $^{11}C_a \times ^5C_b \times ^4C_c$, $a+b+c=6$ |
| $4V + 2C$: $^{11}C_4 \times ^5C_2$; $5V + 1C$: $^{11}C_5 \times ^5C_1$ | B1 | 2 correct unsimplified options |
| $6600 + 3300 + 2310$ | M1 | Add 2 or 3 correct scenarios only |
| $= 12210$ | A1 | Correct answer |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4! \times 3!$ | M1 | $k$ multiplied by $3!$ or $4!$, $k$ an integer $\geq 1$ |
| | A1 | Correct unsimplified expression |
| $= 144$ | A1 | Correct answer |3 In an orchestra, there are 11 violinists, 5 cellists and 4 double bass players. A small group of 6 musicians is to be selected from these 20.\\
(i) How many different selections of 6 musicians can be made if there must be at least 4 violinists, at least 1 cellist and no more than 1 double bass player?\\
The small group that is selected contains 4 violinists, 1 cellist and 1 double bass player. They sit in a line to perform a concert.\\[0pt]
(ii) How many different arrangements are there of these 6 musicians if the violinists must sit together? [3]\\
\hfill \mbox{\textit{CAIE S1 2018 Q3 [7]}}