| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Combinatorial selection with category constraints |
| Difficulty | Moderate -0.8 Part (i) is a standard 'treat as a block' arrangement problem (9! × 3! = 2,177,280). Part (ii) is straightforward application of combination formula C(6,2) × C(3,2) × C(2,1) = 90. Both parts require only direct recall of textbook techniques with no problem-solving insight, making this easier than average but not trivial due to the calculation steps involved. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3! \times 8! \times 9\) | M1 | For \(k3!\) seen, \(k\) a +ve integer, accept \(_3P_3\) |
| M1 | For using \(m8!\) or \(n9!\) seen, \(m\) and \(n\) +ve integers, accept \(m\,{}_8P_8\) etc | |
| \(= 2{,}177{,}280\) or \(2{,}180{,}000\) | A1 3 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(_6C_2 \times {}_3C_2 \times {}_2C_1\) | M1 | Multiplying 3 combinations or 3 numbers or 3 permutations together only |
| B1 | All of \(_6C_2\) and \(_3C_2\) and \(_2C_1\) seen (15, 3, 2) | |
| \(= 90\) | A1 3 | Correct answer |
## Question 3:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $3! \times 8! \times 9$ | M1 | For $k3!$ seen, $k$ a +ve integer, accept $_3P_3$ |
| | M1 | For using $m8!$ or $n9!$ seen, $m$ and $n$ +ve integers, accept $m\,{}_8P_8$ etc |
| $= 2{,}177{,}280$ or $2{,}180{,}000$ | A1 **3** | Correct final answer |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $_6C_2 \times {}_3C_2 \times {}_2C_1$ | M1 | Multiplying 3 combinations or 3 numbers or 3 permutations together only |
| | B1 | All of $_6C_2$ and $_3C_2$ and $_2C_1$ seen (15, 3, 2) |
| $= 90$ | A1 **3** | Correct answer |
---3 Issam has 11 different CDs, of which 6 are pop music, 3 are jazz and 2 are classical.\\
(i) How many different arrangements of all 11 CDs on a shelf are there if the jazz CDs are all next to each other?\\
(ii) Issam makes a selection of 2 pop music CDs, 2 jazz CDs and 1 classical CD. How many different possible selections can be made?
\hfill \mbox{\textit{CAIE S1 2008 Q3 [6]}}