| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with alternating patterns |
| Difficulty | Challenging +1.2 This is a multi-part permutations question requiring systematic case-work and careful counting. Part (a) needs alternating pattern reasoning (placing consonants in gaps between vowels), part (b) uses complementary counting with restrictions, and part (c) involves combinations with 'at least one' probability. While methodical, these are standard S1 techniques without requiring novel insight, making it moderately above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{5! \times 4!}{2! \times 2!}\) | M1 | \(\frac{5! \times 4!}{e}\), \(e\) a positive integer, 1 can be implied. No other terms on numerator. No addition etc. |
| M1 | \(\frac{f}{2! \times g!}\), \(f\) a positive integer, \(g = 1, 2\). No other terms on denominator. | |
| \(720\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{7!}{2!} - 6!\) | B1 | \(\frac{7!}{2!} - e\), \(^7P_5 - e\), \(e\) a positive integer or 0. |
| M1 | \(d - \frac{6!}{r!}\), \(d > 720\), \(r = 1, 2\) | |
| \(= 1800\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5! \times \frac{^6P_2}{2!}\) or \(5! \times \frac{6 \times 5}{2}\) or \(5! \times {^6C_2}\) | B1 | \(5! \times s\), \(s\) a positive integer, 1 may be implied. |
| M1 | \(t \times \frac{6 \times 5}{u}\), \(t\) a positive integer \(> 1\), \(u = 1, 2\) | |
| \(= 1800\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Scenario | Value | Mark |
| \(AD\wedge\wedge\wedge\wedge\wedge A\) | \(5! \times 5 = 600\) | B1 |
| \(A\wedge D\wedge\wedge\wedge\wedge A\) | \(5! \times 4 = 480\) | M1 |
| \(A\wedge\wedge D\wedge\wedge\wedge A\) | \(5! \times 3 = 360\) | |
| \(A\wedge\wedge\wedge D\wedge\wedge A\) | \(5! \times 2 = 240\) | |
| \(A\wedge\wedge\wedge\wedge D\wedge A\) | \(5! \times 1 = 120\) | |
| Total \(= 1800\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Scenario \(AD\wedge\wedge\): \(^2C_1 \times {^2C_1} \times {^5C_2} = 40\) | M1 | At least one correct unsimplified expression for an identified scenario. |
| Scenario \(ADD\wedge\): \(^2C_1 \times [{^2C_2} \times] {^5C_1} = 10\) | ||
| Total \(= 40 + 10 = 50\) | A1 | www; If M0 scored, SC B1 [total \(=\)] 50 www. |
| Total number of selections \(= {^9C_4} = 126\) | B1 | Accept evaluated, accept as denominator of probability expression. Do not condone \(^9C_5\) unless clear explanation for selecting letters not in the group. |
| Probability \(= \frac{50}{126} = \frac{25}{63}\) | B1 FT | \(0.396825\ldots\) to at least 3SF. FT \(\frac{\text{their attempted } 40+10}{126}\). Numerator must be from attempt to find the 2 appropriate scenarios and must be evaluated. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AD\wedge\wedge\): \(\frac{2}{9} \times \frac{2}{8} \times \frac{5}{7} \times \frac{4}{6} \times {^4P_2} = \frac{960}{3024} = \frac{20}{63}\) | M1 | Numerator for at least one correct unsimplified expression for an identified scenario: either \(\frac{2\times2\times5\times4\times12}{a\times b\times c\times d}\) or \(\frac{2\times2\times1\times5\times12}{a\times b\times c\times d}\), \(6 \leq a,b,c,d \leq 9\) |
| \(ADD\wedge\): \(\frac{2}{9} \times \frac{2}{8} \times \frac{1}{7} \times \frac{5}{6} \times \frac{^4P_3}{2!} = \frac{240}{3024} = \frac{5}{63}\) | A1 | \(\frac{2\times2\times5\times4\times12}{a\times b\times c\times d} + \frac{2\times2\times1\times5\times12}{a\times b\times c\times d}\), \(6 \leq a,b,c,d \leq 9\). If M0 scored, SC B1 \(\frac{1200}{g}\), \(g > 1200\), or \(\frac{25}{63}\) seen. |
| Total Probability \(= \frac{20}{63} + \frac{5}{63}\) | B1 | \(\frac{p}{9} \times \frac{q}{8} \times \frac{r}{7} \times \frac{s}{6}\) present in all scenarios attempted, accept \(\frac{t}{3024}\), \(t < 3024\). |
| \(\frac{1200}{3024}, \frac{25}{63}\) | B1 FT | \(0.396825\ldots\) to at least 3SF. FT \(\frac{\text{their attempted } 960+240}{3024}\). Numerator must be from attempt to find the 2 appropriate scenarios. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^2C_1 \times (^7C_3 - {^5C_3}) [= 2 \times (35-10)]\) | M1 | \(a \times ({^7C_3} - {^5C_3})\), \(a = 1, 2\) |
| Total \(= 50\) | A1 | www; If M0 scored, SC B1 [total \(=\)] 50 www. |
| Total number of selections \(= {^9C_4} = 126\) | B1 | Accept evaluated; do not condone \(^9C_5\) unless clear explanation. |
| Probability \(= \frac{50}{126}, \frac{25}{63}\) | B1 FT | \(0.396825\ldots\) to at least 3SF. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either 10 correct outcomes for \(ADD\wedge\) listed or 40 correct outcomes for \(AD\wedge\wedge\) listed | M1 | |
| 50 stated | A1 | www; If M0 scored, SC B1 [total \(=\)] 50 www. |
| 126 stated or correct outcomes listed | B1 | |
| Probability \(= \frac{50}{126}, \frac{25}{63}\) | B1 | \(0.396825\ldots\) to at least 3SF. |
| 4 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5! \times 4!}{2! \times 2!}$ | M1 | $\frac{5! \times 4!}{e}$, $e$ a positive integer, 1 can be implied. No other terms on numerator. No addition etc. |
| | M1 | $\frac{f}{2! \times g!}$, $f$ a positive integer, $g = 1, 2$. No other terms on denominator. |
| $720$ | A1 | |
---
## Question 7(b):
**Method 1:** Number of arrangements with A at each end − Number of arrangements with A at each end and 2 Ds together.
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{7!}{2!} - 6!$ | B1 | $\frac{7!}{2!} - e$, $^7P_5 - e$, $e$ a positive integer or 0. |
| | M1 | $d - \frac{6!}{r!}$, $d > 720$, $r = 1, 2$ |
| $= 1800$ | A1 | |
**Method 2:** A $\wedge\wedge\wedge\wedge\wedge$ A and Ds inserted separately
| Answer | Mark | Guidance |
|--------|------|----------|
| $5! \times \frac{^6P_2}{2!}$ or $5! \times \frac{6 \times 5}{2}$ or $5! \times {^6C_2}$ | B1 | $5! \times s$, $s$ a positive integer, 1 may be implied. |
| | M1 | $t \times \frac{6 \times 5}{u}$, $t$ a positive integer $> 1$, $u = 1, 2$ |
| $= 1800$ | A1 | |
**Method 3:** Scenarios with As at each end and Ds placed in different positions.
| Scenario | Value | Mark | Guidance |
|----------|-------|------|----------|
| $AD\wedge\wedge\wedge\wedge\wedge A$ | $5! \times 5 = 600$ | B1 | Correct outcome/value for 1 identified scenario, accept unsimplified. |
| $A\wedge D\wedge\wedge\wedge\wedge A$ | $5! \times 4 = 480$ | M1 | Add values of 5 correct scenarios, no incorrect/repeated scenarios. |
| $A\wedge\wedge D\wedge\wedge\wedge A$ | $5! \times 3 = 360$ | | |
| $A\wedge\wedge\wedge D\wedge\wedge A$ | $5! \times 2 = 240$ | | |
| $A\wedge\wedge\wedge\wedge D\wedge A$ | $5! \times 1 = 120$ | | |
| Total $= 1800$ | | A1 | |
---
## Question 7(c):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Scenario $AD\wedge\wedge$: $^2C_1 \times {^2C_1} \times {^5C_2} = 40$ | M1 | At least one correct unsimplified expression for an identified scenario. |
| Scenario $ADD\wedge$: $^2C_1 \times [{^2C_2} \times] {^5C_1} = 10$ | | |
| Total $= 40 + 10 = 50$ | A1 | www; If M0 scored, SC B1 [total $=$] 50 www. |
| Total number of selections $= {^9C_4} = 126$ | B1 | Accept evaluated, accept as denominator of probability expression. Do not condone $^9C_5$ unless clear explanation for selecting letters not in the group. |
| Probability $= \frac{50}{126} = \frac{25}{63}$ | B1 FT | $0.396825\ldots$ to at least 3SF. FT $\frac{\text{their attempted } 40+10}{126}$. Numerator must be from attempt to find the 2 appropriate scenarios and must be evaluated. |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $AD\wedge\wedge$: $\frac{2}{9} \times \frac{2}{8} \times \frac{5}{7} \times \frac{4}{6} \times {^4P_2} = \frac{960}{3024} = \frac{20}{63}$ | M1 | Numerator for at least one correct unsimplified expression for an identified scenario: either $\frac{2\times2\times5\times4\times12}{a\times b\times c\times d}$ or $\frac{2\times2\times1\times5\times12}{a\times b\times c\times d}$, $6 \leq a,b,c,d \leq 9$ |
| $ADD\wedge$: $\frac{2}{9} \times \frac{2}{8} \times \frac{1}{7} \times \frac{5}{6} \times \frac{^4P_3}{2!} = \frac{240}{3024} = \frac{5}{63}$ | A1 | $\frac{2\times2\times5\times4\times12}{a\times b\times c\times d} + \frac{2\times2\times1\times5\times12}{a\times b\times c\times d}$, $6 \leq a,b,c,d \leq 9$. If M0 scored, SC B1 $\frac{1200}{g}$, $g > 1200$, or $\frac{25}{63}$ seen. |
| Total Probability $= \frac{20}{63} + \frac{5}{63}$ | B1 | $\frac{p}{9} \times \frac{q}{8} \times \frac{r}{7} \times \frac{s}{6}$ present in all scenarios attempted, accept $\frac{t}{3024}$, $t < 3024$. |
| $\frac{1200}{3024}, \frac{25}{63}$ | B1 FT | $0.396825\ldots$ to at least 3SF. FT $\frac{\text{their attempted } 960+240}{3024}$. Numerator must be from attempt to find the 2 appropriate scenarios. |
**Method 3:** Selecting the A and then selecting 3 any letters and removing selections without Ds.
| Answer | Mark | Guidance |
|--------|------|----------|
| $^2C_1 \times (^7C_3 - {^5C_3}) [= 2 \times (35-10)]$ | M1 | $a \times ({^7C_3} - {^5C_3})$, $a = 1, 2$ |
| Total $= 50$ | A1 | www; If M0 scored, SC B1 [total $=$] 50 www. |
| Total number of selections $= {^9C_4} = 126$ | B1 | Accept evaluated; do not condone $^9C_5$ unless clear explanation. |
| Probability $= \frac{50}{126}, \frac{25}{63}$ | B1 FT | $0.396825\ldots$ to at least 3SF. |
**Method 4:** Listing outcomes.
| Answer | Mark | Guidance |
|--------|------|----------|
| Either 10 correct outcomes for $ADD\wedge$ listed or 40 correct outcomes for $AD\wedge\wedge$ listed | M1 | |
| 50 stated | A1 | www; If M0 scored, SC B1 [total $=$] 50 www. |
| 126 stated or correct outcomes listed | B1 | |
| Probability $= \frac{50}{126}, \frac{25}{63}$ | B1 | $0.396825\ldots$ to at least 3SF. |
| | **4** | |7
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements of the 9 letters in the word ANDROMEDA in which no consonant is next to another consonant. (The letters D, M, N and R are consonants and the letters A, E and O are not consonants.)
\item Find the number of different arrangements of the 9 letters in the word ANDROMEDA in which there is an A at each end and the Ds are not together.\\
Four letters are selected at random from the 9 letters in the word ANDROMEDA.
\item Find the probability that this selection contains at least one D and exactly one A .\\
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 2023 Q7 [10]}}