| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with identical objects |
| Difficulty | Moderate -0.3 This is a standard permutations and combinations question covering arrangements with and without identical objects. Parts (i)-(ii) are straightforward applications of P(n,r) and division by factorials for identical items. Parts (iii)-(v) require systematic case analysis but use only basic counting principles taught in S1. Slightly easier than average A-level due to being purely procedural with no novel insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \({^6}P_4 = 6!/2! = 360\) | B1 | Correct answer |
| [1] | ||
| (ii) \(4!/2! = 12\) | B1 | Correct answer |
| [1] | ||
| (iii) \(4! \times {^3}C_3 = 360 \text{ or } {^6}P_4\) | B1 | Correct final answer |
| [1] | ||
| (iv) e.g. \(2R \text{ } 1B \text{ } 1G, \text{ } 1R \text{ } 2B \text{ } 1G, \text{ } 1R \text{ } 1B \text{ } 2G\) | M1 | 4!/2! seen |
| \(\frac{4!}{2!} + \frac{4!}{2!} + \frac{4!}{2!} = 36, \text{ mult by } {^3}C_3\) | M1 | Mult by \(^nC_3\) |
| \(\text{total} = 720\) | A1 | Correct answer |
| [3] | ||
| (v) \(2R \text{ } 2B = 4!/2!/2! = 6\) | M1 | Considering 2 colours e.g. RRBB or RBBR or... |
| \(\text{Mult by } {^6}C_2, \text{ total} = 90\) | A1 | mult by \(^nC_2\) |
| \(\text{Answer} = 360 + 720 + 90 = 1170\) | A1ft | Ft their (iii) + (iv) + (v) |
| [3] |
**(i)** ${^6}P_4 = 6!/2! = 360$ | B1 | Correct answer
| [1] |
**(ii)** $4!/2! = 12$ | B1 | Correct answer
| [1] |
**(iii)** $4! \times {^3}C_3 = 360 \text{ or } {^6}P_4$ | B1 | Correct final answer
| [1] |
**(iv)** e.g. $2R \text{ } 1B \text{ } 1G, \text{ } 1R \text{ } 2B \text{ } 1G, \text{ } 1R \text{ } 1B \text{ } 2G$ | M1 | 4!/2! seen
$\frac{4!}{2!} + \frac{4!}{2!} + \frac{4!}{2!} = 36, \text{ mult by } {^3}C_3$ | M1 | Mult by $^nC_3$
$\text{total} = 720$ | A1 | Correct answer
| [3] |
**(v)** $2R \text{ } 2B = 4!/2!/2! = 6$ | M1 | Considering 2 colours e.g. RRBB or RBBR or...
$\text{Mult by } {^6}C_2, \text{ total} = 90$ | A1 | mult by $^nC_2$
$\text{Answer} = 360 + 720 + 90 = 1170$ | A1ft | Ft their (iii) + (iv) + (v)
| [3] |6\\
\includegraphics[max width=\textwidth, alt={}, center]{fcf7b1c6-cc76-4c84-998c-9de6a7e9bb2d-3_163_618_260_765}
Pegs are to be placed in the four holes shown, one in each hole. The pegs come in different colours and pegs of the same colour are identical. Calculate how many different arrangements of coloured pegs in the four holes can be made using\\
(i) 6 pegs, all of different colours,\\
(ii) 4 pegs consisting of 2 blue pegs, 1 orange peg and 1 yellow peg.
Beryl has 12 pegs consisting of 2 red, 2 blue, 2 green, 2 orange, 2 yellow and 2 black pegs. Calculate how many different arrangements of coloured pegs in the 4 holes Beryl can make using\\
(iii) 4 different colours,\\
(iv) 3 different colours,\\
(v) any of her 12 pegs.
\hfill \mbox{\textit{CAIE S1 2010 Q6 [9]}}