| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Items NOT together (general separation) |
| Difficulty | Moderate -0.8 This is a straightforward permutations question testing basic counting principles. Part (i) is trivial (9!), part (ii) uses standard complement counting (total minus adjacent arrangements), and parts (iii)-(v) apply basic permutation formulas P(9,3) with simple restrictions. All techniques are routine for S1 level with no novel problem-solving required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) 362880 (363000) | B1 [1] | |
| (ii) PG or GP in 8! × 2 = 80640 or 7/9 of (i) | M1 | Considering together and also subtracting from their (i) or using probabilities |
| 362880 – 80640 = 282240 | B1, A1ft [3] | 8! × 2 or 80640 seen oe. correct answer ft 40320 only |
| (iii) \(^3P_3\) or \(^C_3 \times 3!\) or 9!/6! | M1 | \(^nP_r\) or \(^nC_s\) seen allow extra multiplication |
| \(= 504\) | A1 [2] | correct final answer |
| (iv) \(^2C_2 \times 3!\) or 504 – \(^3C_3 \times 3!\) or \(^nP_2 \times 3\) | M1 | \(^nC_r\) or \(^nP_s\), seen allow extra mult, or (iii)/9 or (iii)/3 |
| \(= 168\) | A1 [2] | correct final answer |
| (v) PG and x in \(7 \times 2 \times 2\) ways = 28 | M1 | x × 2 × 2 seen or their (iii) – 7 or \(^1C_1\) or \(^C_2\) |
| Answer 504 – 28 = 476 | A1 [2] | correct answer |
(i) 362880 (363000) | B1 [1] |
(ii) PG or GP in 8! × 2 = 80640 or 7/9 of (i) | M1 | Considering together and also subtracting from their (i) or using probabilities
362880 – 80640 = 282240 | B1, A1ft [3] | 8! × 2 or 80640 seen oe. correct answer ft 40320 only
(iii) $^3P_3$ or $^C_3 \times 3!$ or 9!/6! | M1 | $^nP_r$ or $^nC_s$ seen allow extra multiplication
$= 504$ | A1 [2] | correct final answer
(iv) $^2C_2 \times 3!$ or 504 – $^3C_3 \times 3!$ or $^nP_2 \times 3$ | M1 | $^nC_r$ or $^nP_s$, seen allow extra mult, or (iii)/9 or (iii)/3
$= 168$ | A1 [2] | correct final answer
(v) PG and x in $7 \times 2 \times 2$ ways = 28 | M1 | x × 2 × 2 seen or their (iii) – 7 or $^1C_1$ or $^C_2$
Answer 504 – 28 = 476 | A1 [2] | correct answerNine cards, each of a different colour, are to be arranged in a line.
\begin{enumerate}[label=(\roman*)]
\item How many different arrangements of the 9 cards are possible? [1]
\end{enumerate}
The 9 cards include a pink card and a green card.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item How many different arrangements do not have the pink card next to the green card? [3]
\end{enumerate}
Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item How many different arrangements in total of 3 cards are possible? [2]
\item How many of the arrangements of 3 cards in part (iii) contain the pink card? [2]
\item How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card? [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2010 Q7 [10]}}