| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| 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 common techniques: treating groups as single units (part a), conditional probability with arrangements (part b), and selections with restrictions (part c). All parts use routine methods taught in S1 with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(7!\) | M1 | \(\frac{7!}{b! \times c!}\ b,c = 1,2\); \(7! \times \frac{2!}{2!} \times \frac{2!}{2!}\) oe, no further terms present |
| \(5040\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5! \times 5 \times 2\) | M1 | \(5! \times d\), \(d\) integer \(> 1\) |
| M1 | \(e! \times f \times g\), \(e = 5,6,7\); \(f = 1,5\); \(g = 1,2\); \(f \neq g\), 1 can be implicit | |
| \(1200\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6! \times 2 - 5! \times 2\) | M1 | \(6! \times 2 - h\), \(h\) an integer \(1 < h < 1440\) |
| M1 | \(k - 5! \times 2\), \(k\) an integer \(k > 240\) | |
| \(1200\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\wedge A \wedge\wedge\wedge\wedge A\quad {}^5P_1 \times {}^1P_1 \times {}^5P_5 \times {}^1P_1 = 600\) | M1 | LL treated as a single unit |
| M1 | ||
| \(1200\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total number of arrangements \(= \frac{9!}{2!2!} = 90720\) | B1 | Accept unsimplified. May be seen as denominator of probability. |
| Probability \(= \frac{1200}{90720} = \frac{5}{378} \approx 0.0132\) | B1 FT | \(\frac{\textit{their } 1200}{\textit{their } 90720}\) unsimplified B1 FT if *their* 1200 and *their* 90720 supported by work in this part. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A\_{\_\_\_}\): \({}^5C_4 = 5\); \(AA\_{\_\_}\): \({}^5C_3 = 10\); \(AL\_{\_\_}\): \({}^5C_3 = 10\); \(AAL\_{\_}\): \({}^5C_2 = 10\) | B1 | 1 correct identified outcome/value for A, AL or AAL scenario; accept unsimplified. \({}^5C_{5-x}\) cannot be used in place of \({}^5C_x\) |
| Add 4 values of appropriate scenarios | M1 | No incorrect scenarios, no repeated scenarios; accept unsimplified; condone use of permutations. |
| \([\text{Total} =]\ 35\) | A1 | Value stated WWW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A \wedge\wedge\wedge\wedge\): \({}^7C_4\) | M1 | \({}^7C_h,\ 3 \leqslant h \leqslant 5\) |
| B1 | \({}^7C_4\) oe, no other terms, scenario identified. | |
| \([\text{Total} =]\ 35\) | A1 | Value stated. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A\wedge\wedge\wedge\wedge = {}^6C_4 = 15\); \(AL\wedge\wedge\wedge = {}^6C_3 = 20\) | B1 | Correct outcome/value for 1 identified scenario; accept unsimplified. WWW |
| Add 2 values of appropriate scenarios | M1 | No incorrect scenarios, no repeated scenarios; accept unsimplified; condone use of permutations. |
| \([\text{Total} =]\ 35\) | A1 | Value stated. |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $7!$ | M1 | $\frac{7!}{b! \times c!}\ b,c = 1,2$; $7! \times \frac{2!}{2!} \times \frac{2!}{2!}$ oe, no further terms present |
| $5040$ | A1 | |
---
## Question 7(b):
**Method 1: Arrangements of 6 letters including Ls between As**
| Answer | Mark | Guidance |
|--------|------|----------|
| $5! \times 5 \times 2$ | M1 | $5! \times d$, $d$ integer $> 1$ |
| | M1 | $e! \times f \times g$, $e = 5,6,7$; $f = 1,5$; $g = 1,2$; $f \neq g$, 1 can be implicit |
| $1200$ | A1 | |
**Method 2: Number of arrangements of $LL^{\wedge\wedge\wedge\wedge\wedge}$ minus arrangements with Ls split by an A**
| Answer | Mark | Guidance |
|--------|------|----------|
| $6! \times 2 - 5! \times 2$ | M1 | $6! \times 2 - h$, $h$ an integer $1 < h < 1440$ |
| | M1 | $k - 5! \times 2$, $k$ an integer $k > 240$ |
| $1200$ | A1 | |
**Method 3: Alternative approaches to Method 1**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\wedge A \wedge\wedge\wedge\wedge A\quad {}^5P_1 \times {}^1P_1 \times {}^5P_5 \times {}^1P_1 = 600$ | M1 | LL treated as a single unit |
| | M1 | |
| $1200$ | A1 | |
# Question 7(b):
**Final 2 marks of Question 7(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Total number of arrangements $= \frac{9!}{2!2!} = 90720$ | B1 | Accept unsimplified. May be seen as denominator of probability. |
| Probability $= \frac{1200}{90720} = \frac{5}{378} \approx 0.0132$ | B1 FT | $\frac{\textit{their } 1200}{\textit{their } 90720}$ unsimplified B1 FT if *their* 1200 and *their* 90720 supported by work in this part. |
**Total: 5 marks** (for whole of 7(b))
---
# Question 7(c):
## Method 1: Scenarios identified — Both As and Ls removed
| Answer | Mark | Guidance |
|--------|------|----------|
| $A\_{\_\_\_}$: ${}^5C_4 = 5$; $AA\_{\_\_}$: ${}^5C_3 = 10$; $AL\_{\_\_}$: ${}^5C_3 = 10$; $AAL\_{\_}$: ${}^5C_2 = 10$ | B1 | 1 correct identified outcome/value for A, AL or AAL scenario; accept unsimplified. ${}^5C_{5-x}$ cannot be used in place of ${}^5C_x$ |
| Add 4 values of appropriate scenarios | M1 | No incorrect scenarios, no repeated scenarios; accept unsimplified; condone use of permutations. |
| $[\text{Total} =]\ 35$ | A1 | Value stated WWW. |
## Method 2: 1 A fixed, 1 L removed — No other scenarios can be present anywhere in solution
| Answer | Mark | Guidance |
|--------|------|----------|
| $A \wedge\wedge\wedge\wedge$: ${}^7C_4$ | M1 | ${}^7C_h,\ 3 \leqslant h \leqslant 5$ |
| | B1 | ${}^7C_4$ oe, no other terms, scenario identified. |
| $[\text{Total} =]\ 35$ | A1 | Value stated. |
## Method 3: 1 A fixed, both Ls removed
| Answer | Mark | Guidance |
|--------|------|----------|
| $A\wedge\wedge\wedge\wedge = {}^6C_4 = 15$; $AL\wedge\wedge\wedge = {}^6C_3 = 20$ | B1 | Correct outcome/value for 1 identified scenario; accept unsimplified. WWW |
| Add 2 values of appropriate scenarios | M1 | No incorrect scenarios, no repeated scenarios; accept unsimplified; condone use of permutations. |
| $[\text{Total} =]\ 35$ | A1 | Value stated. |
**Total: 3 marks**7
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements of the 9 letters in the word ALLIGATOR in which the two As are together and the two Ls are together.
\item The 9 letters in the word ALLIGATOR are arranged in a random order.
Find the probability that the two Ls are together and there are exactly 6 letters between the two As.
\item Find the number of different selections of 5 letters from the 9 letters in the word ALLIGATOR which contain at least one A and at most one L.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q7 [10]}}