| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| 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 question with straightforward constraints. Part (i) requires counting cases (2M4W, 1M5W, 0M6W), part (ii) uses complementary counting (total minus forbidden case), and part (iii) is basic permutations with a position restriction. All techniques are routine for S1 level with no novel insight required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(W(8)\) \(M(5)\): \(4 = ^5C_4 \times ^3C_1 = 700\); \(5 = 1 = ^5C_5 \times ^3C_1 = 280\); \(6 = 0 = ^5C_6 \times ^3C_0 = 28\); Total = 1008 | M1 M1 A1 A1 | Mult 2 combs, \(^5C_x \times ^3C_y\); Summing 2 or 3 options; 2 correct options unsimplified; Correct answer |
| 4 marks | ||
| (ii) \(M1\) and \(MMWWW = ^5C_2 \times ^5C_3 = 168\); \(M2\) and \(MMWWW = ^5C_2 \times ^5C_3 = 168\); Neither and \(MMWWWW = ^5C_1 \times ^5C_3 = 56\); Total = 392 | M1 B1 A1 | Summing 3 options; One correct option; Correct answer |
| 3 marks | ||
| OR total, no restrictions = \(^5C_3 \times ^5C_3 = 560\); \(M1M2\) and \(MWWW = ^5C_1 \times ^5C_3 = 168\); \(560 - 168 = 392\) | M1 B1 A1 | Subt 2 men together from no restrictions; One correct of 560 or 168; Correct answer |
| (iii) e.g. \(WWMWMW = 5!\) (women) \(\times 4 = 480\) | M1 M1 A1 | 5! Seen mult by integer ⩾ 1; Mult by 4; Correct answer |
| 3 marks | ||
| OR \(6! - MWWWWW - WWWWWM = 6! - 5! - 5! = 480\) | M1 M1 A1 | 6! seen with a subtraction; 5! or 2 × 5! Seen subtracted; Correct answer |
**(i)** $W(8)$ $M(5)$: $4 = ^5C_4 \times ^3C_1 = 700$; $5 = 1 = ^5C_5 \times ^3C_1 = 280$; $6 = 0 = ^5C_6 \times ^3C_0 = 28$; Total = 1008 | M1 M1 A1 A1 | Mult 2 combs, $^5C_x \times ^3C_y$; Summing 2 or 3 options; 2 correct options unsimplified; Correct answer |
| | | 4 marks |
**(ii)** $M1$ and $MMWWW = ^5C_2 \times ^5C_3 = 168$; $M2$ and $MMWWW = ^5C_2 \times ^5C_3 = 168$; Neither and $MMWWWW = ^5C_1 \times ^5C_3 = 56$; Total = 392 | M1 B1 A1 | Summing 3 options; One correct option; Correct answer |
| | | 3 marks |
| OR total, no restrictions = $^5C_3 \times ^5C_3 = 560$; $M1M2$ and $MWWW = ^5C_1 \times ^5C_3 = 168$; $560 - 168 = 392$ | M1 B1 A1 | Subt 2 men together from no restrictions; One correct of 560 or 168; Correct answer |
**(iii)** e.g. $WWMWMW = 5!$ (women) $\times 4 = 480$ | M1 M1 A1 | 5! Seen mult by integer ⩾ 1; Mult by 4; Correct answer |
| | | 3 marks |
| OR $6! - MWWWWW - WWWWWM = 6! - 5! - 5! = 480$ | M1 M1 A1 | 6! seen with a subtraction; 5! or 2 × 5! Seen subtracted; Correct answer |7 A committee of 6 people is to be chosen from 5 men and 8 women. In how many ways can this be done\\
(i) if there are more women than men on the committee,\\
(ii) if the committee consists of 3 men and 3 women but two particular men refuse to be on the committee together?
One particular committee consists of 5 women and 1 man.\\
(iii) In how many different ways can the committee members be arranged in a line if the man is not at either end?
\hfill \mbox{\textit{CAIE S1 2014 Q7 [10]}}