| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with grouped categories |
| Difficulty | Standard +0.8 This is a multi-part combinatorics problem requiring careful case analysis in part (i), grouped permutations in part (ii), and a challenging derangement-style constraint in part (iii). Part (iii) particularly requires systematic thinking about placing non-adjacent items, which is conceptually harder than standard arrangements and goes beyond routine textbook exercises. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| H=1, J=28, O=2: \({}_{4}C2 \times {}_{9}C8 \times {}_{2}C2 = 54\) | M1 | Mult 3 combs, \({}_{2}C2\) may be implied; \(4Cx \times 9Cy \times 2Cz\) |
| H=3, J=7, O=2: \({}_{4}C3 \times {}_{9}C7 \times {}_{2}C2 = 144\) | M1 | Summing 2 or 3 three-factor options |
| H=4, J=6, O=2: \({}_{4}C4 \times {}_{9}C6 \times {}_{2}C2 = 84\) | A1 | 2 options correct unsimplified |
| Total \(= 282\) ways | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4! \times 6! \times 2! \times 3!\) | M1 | \(4! \times 6! \times 2!\) oe seen multiplied by int \(\geq 1\) |
| M1 | \(3!\) seen multiplied by int \(\geq 1\) | |
| \(= 207360\ (207000)\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 8 J and O trees in \(8! = 40320\) ways | B1 | \(8!\) seen multiplied by int \(\geq 1\), no division |
| 9 gaps \(\times\ 8 \times 7 \times 6\) | M1 | \(9P4\) oe or \(7P4\) or \(8P4\) seen multiplied by int \(\geq 1\), no division |
| \(= 121{,}927{,}680\ (122{,}000{,}000)\) | A1 [3] | Correct answer |
| Answer | Marks |
|---|---|
| SR | Marks |
| (i) SR \({}_{4}C2 \times {}_{9}C2 \times {}_{2}C2 \times {}_{9}C6\) | M1 |
| (ii) SR \(\frac{4!\times6!\times2!}{4!\times6!\times2!}\) or \(3!\) or both M1 | M1 |
| (iii) SR1 \(12! - 9!\ 4!\) | M1 |
| (iii) SR2 \(\frac{9P4}{4!}\) or \(\frac{8!}{6!\ 2!}\) or both | M1 |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| H=1, J=28, O=2: ${}_{4}C2 \times {}_{9}C8 \times {}_{2}C2 = 54$ | M1 | Mult 3 combs, ${}_{2}C2$ may be implied; $4Cx \times 9Cy \times 2Cz$ |
| H=3, J=7, O=2: ${}_{4}C3 \times {}_{9}C7 \times {}_{2}C2 = 144$ | M1 | Summing 2 or 3 three-factor options |
| H=4, J=6, O=2: ${}_{4}C4 \times {}_{9}C6 \times {}_{2}C2 = 84$ | A1 | 2 options correct unsimplified |
| Total $= 282$ ways | A1 **[4]** | Correct answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4! \times 6! \times 2! \times 3!$ | M1 | $4! \times 6! \times 2!$ oe seen multiplied by int $\geq 1$ |
| | M1 | $3!$ seen multiplied by int $\geq 1$ |
| $= 207360\ (207000)$ | A1 **[3]** | Correct answer |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| 8 J and O trees in $8! = 40320$ ways | B1 | $8!$ seen multiplied by int $\geq 1$, no division |
| 9 gaps $\times\ 8 \times 7 \times 6$ | M1 | $9P4$ oe or $7P4$ or $8P4$ seen multiplied by int $\geq 1$, no division |
| $= 121{,}927{,}680\ (122{,}000{,}000)$ | A1 **[3]** | Correct answer |
**Special Rulings:**
| SR | Marks |
|---|---|
| (i) SR ${}_{4}C2 \times {}_{9}C2 \times {}_{2}C2 \times {}_{9}C6$ | M1 |
| (ii) SR $\frac{4!\times6!\times2!}{4!\times6!\times2!}$ or $3!$ or both M1 | M1 |
| (iii) SR1 $12! - 9!\ 4!$ | M1 |
| (iii) SR2 $\frac{9P4}{4!}$ or $\frac{8!}{6!\ 2!}$ or both | M1 |
---6 A town council plans to plant 12 trees along the centre of a main road. The council buys the trees from a garden centre which has 4 different hibiscus trees, 9 different jacaranda trees and 2 different oleander trees for sale.\\
(i) How many different selections of 12 trees can be made if there must be at least 2 of each type of tree?
The council buys 4 hibiscus trees, 6 jacaranda trees and 2 oleander trees.\\
(ii) How many different arrangements of these 12 trees can be made if the hibiscus trees have to be next to each other, the jacaranda trees have to be next to each other and the oleander trees have to be next to each other?\\
(iii) How many different arrangements of these 12 trees can be made if no hibiscus tree is next to another hibiscus tree?
\hfill \mbox{\textit{CAIE S1 2013 Q6 [10]}}