| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Committee with gender/category constraints |
| Difficulty | Moderate -0.8 This is a straightforward combinations question with standard constraints. Parts (i) and (ii) are routine applications of C(n,r) with simple case-by-case counting. Part (iii) adds a mild exclusion principle twist but follows a standard pattern (total minus restricted cases). The calculations are mechanical with no conceptual depth or problem-solving insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \({}_{6}C_3 \times {}_{4}C_2 = 120\) | M1 | For multiplying 2 combinations together, not adding, no perms; \({}_{10}C_3 \times {}_{10}C_2\) or \({}_{5}C_3 \times {}_{5}C_2\) would get M1 |
| \(= 120\) | A1 (2) | For answer 120 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \({}_{6}C_4 \times {}_{4}C_1 (= 60)\) | M1 | For reasonable attempt on option 4M 1W, or 5M 0W, can have \(+\) here and perms |
| \({}_{6}C_5 \times {}_{4}C_0 (= 6)\) | M1 | For other option attempt |
| Answer \(= 186\) | A1 (3) | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Man and woman both on: \({}_{5}C_2 \times {}_{3}C_1 (= 30)\) | M1 | For finding number of ways of man and woman being on together; need not be evaluated but must be multiplied |
| \(120 - 30 = 90\) | M1 | For subtracting a relevant number from their (i) |
| \(= 90\) | A1 (3) | For correct answer |
| OR \({}_{5}C_2 \times {}_{3}C_2 (= 30)\) / \({}_{3}C_1 \times {}_{5}C_3 (= 30)\) / \({}_{5}C_3 \times {}_{3}C_2 (= 30)\); \(\sum = 90\) | M1/M1/A1 (3) | Any 2 of man in woman out / woman in man out / neither in |
| OR \({}_{3}C_1 \times {}_{5}C_3 (= 30)\) / \({}_{3}C_2 \times {}_{6}C_3 (= 60)\); \(\sum = 90\) | M1/M1/A1 (3) | Woman in man out / woman out any man |
| OR \({}_{5}C_2 \times {}_{3}C_2 (= 30)\) / \({}_{5}C_3 \times {}_{4}C_2 (= 60)\); \(\sum = 90\) | M1/M1/A1 (3) | Man in woman out / man out any woman |
## Question 5:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| ${}_{6}C_3 \times {}_{4}C_2 = 120$ | M1 | For multiplying 2 combinations together, not adding, no perms; ${}_{10}C_3 \times {}_{10}C_2$ or ${}_{5}C_3 \times {}_{5}C_2$ would get M1 |
| $= 120$ | A1 (2) | For answer 120 |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| ${}_{6}C_4 \times {}_{4}C_1 (= 60)$ | M1 | For reasonable attempt on option 4M 1W, or 5M 0W, can have $+$ here and perms |
| ${}_{6}C_5 \times {}_{4}C_0 (= 6)$ | M1 | For other option attempt |
| Answer $= 186$ | A1 (3) | For correct answer |
**(iii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Man and woman both on: ${}_{5}C_2 \times {}_{3}C_1 (= 30)$ | M1 | For finding number of ways of man and woman being on together; need not be evaluated but must be multiplied |
| $120 - 30 = 90$ | M1 | For subtracting a relevant number from their **(i)** |
| $= 90$ | A1 (3) | For correct answer |
| **OR** ${}_{5}C_2 \times {}_{3}C_2 (= 30)$ / ${}_{3}C_1 \times {}_{5}C_3 (= 30)$ / ${}_{5}C_3 \times {}_{3}C_2 (= 30)$; $\sum = 90$ | M1/M1/A1 (3) | Any 2 of man in woman out / woman in man out / neither in |
| **OR** ${}_{3}C_1 \times {}_{5}C_3 (= 30)$ / ${}_{3}C_2 \times {}_{6}C_3 (= 60)$; $\sum = 90$ | M1/M1/A1 (3) | Woman in man out / woman out any man |
| **OR** ${}_{5}C_2 \times {}_{3}C_2 (= 30)$ / ${}_{5}C_3 \times {}_{4}C_2 (= 60)$; $\sum = 90$ | M1/M1/A1 (3) | Man in woman out / man out any woman |
---5 A committee of 5 people is to be chosen from 6 men and 4 women. In how many ways can this be done\\
(i) if there must be 3 men and 2 women on the committee,\\
(ii) if there must be more men than women on the committee,\\
(iii) if there must be 3 men and 2 women, and one particular woman refuses to be on the committee with one particular man?
\hfill \mbox{\textit{CAIE S1 2003 Q5 [8]}}