| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with couples/pairs |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question testing standard techniques: treating pairs as units, dividing groups, and systematic counting with constraints. All parts require only direct application of formulas (7!×2^7, 2×7!×7!, basic combinations) with no novel problem-solving or insight needed. Easier than average A-level due to routine nature. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) 7 couples in 7! ways | B1 | 7! seen multiplied |
| each couple in 2 ways so \(7! \times 2^7 = 645120\) | M1 | mult by \(2^7\) |
| A1 [3] | correct final answer | |
| OR \(14 \times 12 \times 10 \times 8 \times 6 \times 4 \times 2 = 645120\) | B2 | correct unsimplified answer |
| A1 | correct answer | |
| (ii) \(7! \times 7! \times 2 = 50,803,200 (50,800,000)\) | B1 | 7! × 7! seen |
| B1 [2] | Correct answer | |
| OR \(14 \times 6! \times 7!\) | B1 | 14 × 7! seen |
| B1 | Correct answer | |
| (b) (i) \(7C2 = 21\) | B1 [1] | |
| (ii) all in: 1 | M1 | Considering both cases |
| all not in: 5C4 = 5 | ||
| total: 6 | A1 [2] | Correct answer |
| (iii) 2 girls in: 6C2 × 3C2 = 45 | M1 | Attempt at summing 2 and 3 girls in the team need not see 3C2 |
| 3 girls in: 6C1 = 6 | ||
| Total: 51 | A1 [2] | Correct answer |
**(a) (i)** 7 couples in 7! ways | B1 | 7! seen multiplied
each couple in 2 ways so $7! \times 2^7 = 645120$ | M1 | mult by $2^7$
| A1 [3] | correct final answer
**OR** $14 \times 12 \times 10 \times 8 \times 6 \times 4 \times 2 = 645120$ | B2 | correct unsimplified answer
| A1 | correct answer
**(ii)** $7! \times 7! \times 2 = 50,803,200 (50,800,000)$ | B1 | 7! × 7! seen
| B1 [2] | Correct answer
**OR** $14 \times 6! \times 7!$ | B1 | 14 × 7! seen
| B1 | Correct answer
**(b) (i)** $7C2 = 21$ | B1 [1]
**(ii)** all in: 1 | M1 | Considering both cases
all not in: 5C4 = 5 |
total: 6 | A1 [2] | Correct answer
**(iii)** 2 girls in: 6C2 × 3C2 = 45 | M1 | Attempt at summing 2 and 3 girls in the team need not see 3C2
3 girls in: 6C1 = 6 |
Total: 51 | A1 [2] | Correct answer7
\begin{enumerate}[label=(\alph*)]
\item Seven friends together with their respective partners all meet up for a meal. To commemorate the occasion they arrange for a photograph to be taken of all 14 of them standing in a line.
\begin{enumerate}[label=(\roman*)]
\item How many different arrangements are there if each friend is standing next to his or her partner?
\item How many different arrangements are there if the 7 friends all stand together and the 7 partners all stand together?
\end{enumerate}\item A group of 9 people consists of 2 boys, 3 girls and 4 adults. In how many ways can a team of 4 be chosen if
\begin{enumerate}[label=(\roman*)]
\item both boys are in the team,
\item the adults are either all in the team or all not in the team,
\item at least 2 girls are in the team?
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2012 Q7 [10]}}