| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | People arrangements in groups/rows |
| Difficulty | Standard +0.8 This is a multi-part combinatorics problem requiring careful case analysis in part (iii) where constraints interact (tenors together, basses together, but 3 specific tenors cannot be adjacent to basses). Parts (i-ii) are standard, but part (iii) requires systematic enumeration of valid arrangements with multiple overlapping restrictions, which is significantly above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(^{13}C_{10} \times ^{12}C_9 \times ^6C_4 \times ^7C_4\) | M1 | Expression involving the product of 4 combinations |
| \(= 33033000\) (33000000) | A1 [2] | Correct final answer; allow \(33 \times 10^6\) or \(3.3 \times 10^7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5! \times 6!\) | B1 | 6! or 5! or 4! oe seen, no denom |
| \(= 86400\) | M1 | A single product involving 6! and either 4! or 5!, no denom |
| A1 [3] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4! \times 3! \times 2\) | B1 | 4! or 3! or \(4!/4\) seen |
| \(= 288\) | M1 | A single product involving 3! (or \(4!/4\)) and 4! |
| A1 [3] | Correct final answer |
## Question 4:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $^{13}C_{10} \times ^{12}C_9 \times ^6C_4 \times ^7C_4$ | M1 | Expression involving the product of 4 combinations |
| $= 33033000$ (33000000) | A1 **[2]** | Correct final answer; allow $33 \times 10^6$ or $3.3 \times 10^7$ |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5! \times 6!$ | B1 | 6! or 5! or 4! oe seen, no denom |
| $= 86400$ | M1 | A single product involving 6! and either 4! or 5!, no denom |
| | A1 **[3]** | Correct final answer |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4! \times 3! \times 2$ | B1 | 4! or 3! or $4!/4$ seen |
| $= 288$ | M1 | A single product involving 3! (or $4!/4$) and 4! |
| | A1 **[3]** | Correct final answer |
---4 A choir consists of 13 sopranos, 12 altos, 6 tenors and 7 basses. A group consisting of 10 sopranos, 9 altos, 4 tenors and 4 basses is to be chosen from the choir.\\
(i) In how many different ways can the group be chosen?\\
(ii) In how many ways can the 10 chosen sopranos be arranged in a line if the 6 tallest stand next to each other?\\
(iii) The 4 tenors and 4 basses in the group stand in a single line with all the tenors next to each other and all the basses next to each other. How many possible arrangements are there if three of the tenors refuse to stand next to any of the basses?
\hfill \mbox{\textit{CAIE S1 2009 Q4 [8]}}