| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| 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 (a) requires treating groups as single units (a common textbook technique), and part (b) uses complementary counting. Both are routine S1 methods with straightforward execution, though the presence of repeated letters adds minor complexity beyond the most basic arrangements questions. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6!\) | M1 | |
| \(720\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total number: \(\frac{9!}{3!2!} = 30240\) | M1 | |
| Number with Ls together \(= \frac{8!}{3!} = 6720\) | M1 | |
| Number with Ls not together \(= \frac{9!}{3!2!} - \frac{8!}{3!} = 30240 - 6720\) | M1 | |
| \(23520\) | A1 | |
| Alternative: \(\frac{7!}{3!} \times \frac{8 \times 7}{2}\) | ||
| \(7! \times k\) in numerator, \(k\) integer \(\geq 1\) | M1 | |
| \(8 \times 7 \times m\) in numerator or \(8C2 \times m\), \(m\) integer \(\geq 1\) | M1 | |
| \(3!\) in denominator | M1 | |
| \(23520\) | A1 |
## Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $6!$ | M1 | |
| $720$ | A1 | |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Total number: $\frac{9!}{3!2!} = 30240$ | M1 | |
| Number with Ls together $= \frac{8!}{3!} = 6720$ | M1 | |
| Number with Ls not together $= \frac{9!}{3!2!} - \frac{8!}{3!} = 30240 - 6720$ | M1 | |
| $23520$ | A1 | |
| **Alternative:** $\frac{7!}{3!} \times \frac{8 \times 7}{2}$ | | |
| $7! \times k$ in numerator, $k$ integer $\geq 1$ | M1 | |
| $8 \times 7 \times m$ in numerator or $8C2 \times m$, $m$ integer $\geq 1$ | M1 | |
| $3!$ in denominator | M1 | |
| $23520$ | A1 | |2
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements that can be made from the 9 letters of the word JEWELLERY in which the three Es are together and the two Ls are together.
\item Find the number of different arrangements that can be made from the 9 letters of the word JEWELLERY in which the two Ls are not next to each other.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q2 [6]}}