| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with adjacency requirements |
| Difficulty | Standard +0.3 This is a standard permutations question with adjacency constraints, slightly above average difficulty. Part (i) uses the classic 'treat as a block' technique (3!×3!=36). Part (ii) requires arranging the remaining two people first then inserting the three (2!×3!×2×1=24). Part (iii) needs careful counting of complementary cases, requiring more problem-solving than routine recall but still a well-established question type for S1. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(3! \times 3! = 36\) | M1 | For at least one factor of \(3!\) |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Ali, Bev and Carla must be in 1st, 3rd, 5th, posns. Hence number of ways is \(3! \times 2! = 12\) | B1 | For identifying this restriction |
| M1 | For at least one of the factors present | |
| A1 | For correct answer |
Total number of possible arrangements is \(5!\)
| Answer | Marks | Guidance |
|---|---|---|
| Hence probability is \(\frac{72}{120} = \frac{3}{5}\) | B1 | For correct statement or use of \(5!\) |
| M1 | For subtraction of (i) and (ii) from total | |
| A1 | For correct answer |
**Part (i)**
$3! \times 3! = 36$ | M1 | For at least one factor of $3!$
| A1 | For correct answer
**Part (ii)**
Ali, Bev and Carla must be in 1st, 3rd, 5th, posns. Hence number of ways is $3! \times 2! = 12$ | B1 | For identifying this restriction
| M1 | For at least one of the factors present
| A1 | For correct answer
**Part (iii)**
Total number of possible arrangements is $5!$
No. of ways with 2 together is $5! - 36 - 12 = 72$
Hence probability is $\frac{72}{120} = \frac{3}{5}$ | B1 | For correct statement or use of $5!$
| M1 | For subtraction of (i) and (ii) from total
| A1 | For correct answer
---3 Five friends, Ali, Bev, Carla, Don and Ed, stand in a line for a photograph.\\
(i) How many different possible arrangements are there if Ali, Bev and Carla stand next to each other?\\
(ii) How many different possible arrangements are there if none of Ali, Bev and Carla stand next to each other?\\
(iii) If all possible arrangements are equally likely, find the probability that two of Ali, Bev and Carla are next to each other, but the third is not next to either of the other two.
\hfill \mbox{\textit{OCR S1 Q3 [8]}}