| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Multiple separation conditions combined |
| Difficulty | Standard +0.3 Part (a) is straightforward permutations with identical objects after fixing positions. Part (b) requires treating blue candles as a block, then using complementary counting to exclude arrangements where red candles are together—a standard two-step technique. Both parts use routine methods with no novel insight required, making this slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R \wedge\wedge\wedge\wedge\wedge\wedge\wedge\wedge R\); \(\frac{9!}{3!6!}\) | M1 | 9! alone on numerator; \(3! \times k\) or \(6! \times k\) on denominator |
| \(= 84\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\wedge(BBB)\wedge\wedge\wedge\wedge\wedge\); \(\frac{7!}{6!} \times \frac{8 \times 7}{2}\) | M1 | \(\frac{7!}{6!} \times k\) or \(7k\) seen, \(k\) an integer \(> 0\) |
| M1 | \(m \times n(n-1)\) or \(m \times {}^nC_2\) or \(m \times {}^nP_2\), \(n = 7, 8\) or \(9\), \(m\) an integer \(> 0\) | |
| M1 | \(n = 8\) used in above expression | |
| \(= 196\) | A1 | |
| Alternative: \(\frac{9!}{2!6!} - \frac{8!}{6!} = [252 - 56]\) | M1 | 9! seen alone or as numerator with subtraction |
| M1 | 8! seen alone or as numerator in a second term and no other terms | |
| M1 | All terms divided by \(6! \times k\), \(k\) an integer | |
| \(= 196\) | A1 | |
| Total: 4 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R \wedge\wedge\wedge\wedge\wedge\wedge\wedge\wedge R$; $\frac{9!}{3!6!}$ | M1 | 9! alone on numerator; $3! \times k$ or $6! \times k$ on denominator |
| $= 84$ | A1 | |
| **Total: 2** | | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\wedge(BBB)\wedge\wedge\wedge\wedge\wedge$; $\frac{7!}{6!} \times \frac{8 \times 7}{2}$ | M1 | $\frac{7!}{6!} \times k$ or $7k$ seen, $k$ an integer $> 0$ |
| | M1 | $m \times n(n-1)$ or $m \times {}^nC_2$ or $m \times {}^nP_2$, $n = 7, 8$ or $9$, $m$ an integer $> 0$ |
| | M1 | $n = 8$ used in above expression |
| $= 196$ | A1 | |
| **Alternative:** $\frac{9!}{2!6!} - \frac{8!}{6!} = [252 - 56]$ | M1 | 9! seen alone or as numerator with subtraction |
| | M1 | 8! seen alone or as numerator in a second term and no other terms |
| | M1 | All terms divided by $6! \times k$, $k$ an integer |
| $= 196$ | A1 | |
| **Total: 4** | | |4 Richard has 3 blue candles, 2 red candles and 6 green candles. The candles are identical apart from their colours. He arranges the 11 candles in a line.
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements of the 11 candles if there is a red candle at each end.
\item Find the number of different arrangements of the 11 candles if all the blue candles are together and the red candles are not together.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q4 [6]}}