| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Committee with gender/category constraints |
| Difficulty | Standard +0.3 This is a standard combinations problem with constraints that requires case-by-case analysis but uses straightforward techniques. Part (i) involves identifying valid gender distributions (4M2W, 5M1W, 6M0W) and summing combinations. Part (ii) uses complementary counting (total minus both men together). Both parts are routine applications of combination formulas with minimal problem-solving insight required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M(8),\ W(4)\): \(4\ \ 2\): \({}^8C_4 \times {}^4C_2 = 420\) ways; \(5\ \ 1\): \({}^8C_5 \times {}^4C_1 = 224\) ways; \(6\ \ 0\): \({}^8C_6 \times {}^4C_0 = 28\) ways | B1 | One unsimplified product correct |
| Summing correct scenarios | M1 | Summing the number of ways for 2 or 3 correct scenarios (can be unsimplified), no incorrect scenarios |
| Total \(672\) ways | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total number of selections \(= {}^{12}C_6 = 924\) (A) | M1 | \({}^{12}C_x\) — (subtraction seen), accept unsimplified |
| Selections with males together \(= {}^{10}C_4 = 210\) (B) | A1 | Correct unsimplified expression |
| Total \(= (A) - (B) = 714\) | A1 | Correct answer |
| Alternative: No males + Only male 1 + Only male 2 \(= {}^{10}C_6 + {}^{10}C_5 + {}^{10}C_5\) | M1 | \({}^{10}C_x + 2 \times {}^{10}C_y,\ x \neq y\) seen, accept unsimplified |
| \(= 210 + 252 + 252\) | A1 | Correct unsimplified expression |
| \(= 714\) | A1 | Correct answer |
| Alternative: Pool without male 1 + Pool without male 2 \(-\) Pool without either male \(= {}^{11}C_6 + {}^{11}C_6 - {}^{10}C_6 = 462 + 462 - 210\) | M1/A1 | \(2 \times {}^{11}C_x - {}^{10}C_x\) |
| \(= 714\) | A1 | Correct answer |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M(8),\ W(4)$: $4\ \ 2$: ${}^8C_4 \times {}^4C_2 = 420$ ways; $5\ \ 1$: ${}^8C_5 \times {}^4C_1 = 224$ ways; $6\ \ 0$: ${}^8C_6 \times {}^4C_0 = 28$ ways | B1 | One unsimplified product correct |
| Summing correct scenarios | M1 | Summing the number of ways for 2 or 3 correct scenarios (can be unsimplified), no incorrect scenarios |
| Total $672$ ways | A1 | Correct answer |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total number of selections $= {}^{12}C_6 = 924$ (A) | M1 | ${}^{12}C_x$ — (subtraction seen), accept unsimplified |
| Selections with males together $= {}^{10}C_4 = 210$ (B) | A1 | Correct unsimplified expression |
| Total $= (A) - (B) = 714$ | A1 | Correct answer |
| **Alternative:** No males + Only male 1 + Only male 2 $= {}^{10}C_6 + {}^{10}C_5 + {}^{10}C_5$ | M1 | ${}^{10}C_x + 2 \times {}^{10}C_y,\ x \neq y$ seen, accept unsimplified |
| $= 210 + 252 + 252$ | A1 | Correct unsimplified expression |
| $= 714$ | A1 | Correct answer |
| **Alternative:** Pool without male 1 + Pool without male 2 $-$ Pool without either male $= {}^{11}C_6 + {}^{11}C_6 - {}^{10}C_6 = 462 + 462 - 210$ | M1/A1 | $2 \times {}^{11}C_x - {}^{10}C_x$ |
| $= 714$ | A1 | Correct answer |
---4 (i) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if there must be at least twice as many men as there are women on the committee.\\
(ii) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if 2 particular men refuse to be on the committee together.\\
\hfill \mbox{\textit{CAIE S1 2019 Q4 [6]}}