| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with grouped categories |
| Difficulty | Standard +0.8 Part (a) is routine permutations with repetition. Part (b) requires careful handling of multiple constraints (grouping Es, spacing Rs exactly 3 apart) which demands systematic case analysis. Part (c) involves alternating patterns with repeated letters requiring careful counting of valid arrangements. The multi-constraint nature and need for structured reasoning elevates this above standard permutation exercises. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{10!}{2!\,4!} = 75600\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4!\times3!\) | M1 | \(4!\) SOI in all terms leading to final answer. Allow 24 if \(4!=24\) is seen. |
| M1 | Either \(3!\) SOI in expression leading to final answer, or at least 6 distinct scenarios identified and added. Condone 3 distinct scenarios \(\times2\). Ignore repeated scenarios. | |
| \(4!\times3!\) | A1 | Fully correct unsimplified expression leading to final answer. |
| \(144\) | B1 | WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Method 1 (denominator from 7(a), no denominator, or incorrect denominator): | ||
| Numerator: \(\frac{5!}{2!}\times\frac{5!}{4!}\times2 = 600\) | B1 | \(\frac{5!}{2!}\) seen (arrangements of consonants). |
| B1 | \(\frac{5!}{4!}\) seen (arrangements of vowels). | |
| M1 | \(\frac{5!}{r}\times\frac{5!}{s}\times2\), \(r=1\) or \(2\), \(s=1,4,4!\) or \(24\). | |
| \(\text{Probability} = \frac{600}{75600}\) | M1 | \(\frac{\text{their }600}{\text{their }(a)}\) or \(\frac{\text{their }600}{75600}\) |
| \(= \frac{1}{126},\ 0.00794\) | A1 | Accept \(\frac{600}{75600}\) OE. |
| Method 2 (denominator \(10!\)): | ||
| \(5!\times5!\times2 = 28800\) | (B1) | \(5!\) seen (arrangements of consonants). |
| (B1) | A second \(5!\) seen (arrangements of vowels). | |
| (M1) | \(5!\times5!\times k\), \(k=1\) or \(2\). | |
| \(\text{Probability} = \frac{28800}{10!}\) | (M1) | |
| \(= \frac{1}{126},\ 0.00794\) | (A1) | Accept \(\frac{600}{75600}\) OE. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{5}{10}\times\frac{5}{9}\times\frac{4}{8}\times\frac{4}{7}\times\frac{3}{6}\times\frac{3}{5}\times\frac{2}{4}\times\frac{2}{3}\times\frac{1}{2}\times\frac{1}{1}\times 2\) | B1 | \(\frac{5}{a}\times\frac{4}{b}\times\frac{3}{c}\times\frac{2}{d}\times\frac{1}{e}\) seen. \(10\geq a > b > c > d > e \geq 1\) (arrangements of consonants). |
| B1 | A second \(\frac{5}{f}\times\frac{4}{g}\times\frac{3}{h}\times\frac{2}{i}\times\frac{1}{j}\) seen. \(10\geq f > g > h > i > j \geq 1\) (arrangements of vowels). | |
| M1 | \(\frac{5}{a}\times\frac{4}{b}\times\frac{3}{c}\times\frac{2}{d}\times\frac{1}{e}\times\frac{5}{f}\times\frac{4}{g}\times\frac{3}{h}\times\frac{2}{i}\times\frac{1}{j}\times k\), \(k = 1\) or \(2\). | |
| M1 | \(\frac{\textit{their}\ 5\times4\times3\times2\times1\times5\times4\times3\times2\times1}{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}\) | |
| \(=\frac{1}{126}\), \(0.00794\) | A1 | Accept \(\frac{28800}{362800}\) OE. |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{10!}{2!\,4!} = 75600$ | B1 | |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4!\times3!$ | M1 | $4!$ SOI in all terms leading to final answer. Allow 24 if $4!=24$ is seen. |
| | M1 | Either $3!$ SOI in expression leading to final answer, or at least 6 distinct scenarios identified and added. Condone 3 distinct scenarios $\times2$. Ignore repeated scenarios. |
| $4!\times3!$ | A1 | Fully correct unsimplified expression leading to final answer. |
| $144$ | B1 | WWW |
---
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Method 1** (denominator from 7(a), no denominator, or incorrect denominator): | | |
| Numerator: $\frac{5!}{2!}\times\frac{5!}{4!}\times2 = 600$ | B1 | $\frac{5!}{2!}$ seen (arrangements of consonants). |
| | B1 | $\frac{5!}{4!}$ seen (arrangements of vowels). |
| | M1 | $\frac{5!}{r}\times\frac{5!}{s}\times2$, $r=1$ or $2$, $s=1,4,4!$ or $24$. |
| $\text{Probability} = \frac{600}{75600}$ | M1 | $\frac{\text{their }600}{\text{their }(a)}$ or $\frac{\text{their }600}{75600}$ |
| $= \frac{1}{126},\ 0.00794$ | A1 | Accept $\frac{600}{75600}$ OE. |
| **Method 2** (denominator $10!$): | | |
| $5!\times5!\times2 = 28800$ | (B1) | $5!$ seen (arrangements of consonants). |
| | (B1) | A second $5!$ seen (arrangements of vowels). |
| | (M1) | $5!\times5!\times k$, $k=1$ or $2$. |
| $\text{Probability} = \frac{28800}{10!}$ | (M1) | |
| $= \frac{1}{126},\ 0.00794$ | (A1) | Accept $\frac{600}{75600}$ OE. |
## Question 7(c):
**Method 3: Using probabilities**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5}{10}\times\frac{5}{9}\times\frac{4}{8}\times\frac{4}{7}\times\frac{3}{6}\times\frac{3}{5}\times\frac{2}{4}\times\frac{2}{3}\times\frac{1}{2}\times\frac{1}{1}\times 2$ | **B1** | $\frac{5}{a}\times\frac{4}{b}\times\frac{3}{c}\times\frac{2}{d}\times\frac{1}{e}$ seen. $10\geq a > b > c > d > e \geq 1$ (arrangements of consonants). |
| | **B1** | A second $\frac{5}{f}\times\frac{4}{g}\times\frac{3}{h}\times\frac{2}{i}\times\frac{1}{j}$ seen. $10\geq f > g > h > i > j \geq 1$ (arrangements of vowels). |
| | **M1** | $\frac{5}{a}\times\frac{4}{b}\times\frac{3}{c}\times\frac{2}{d}\times\frac{1}{e}\times\frac{5}{f}\times\frac{4}{g}\times\frac{3}{h}\times\frac{2}{i}\times\frac{1}{j}\times k$, $k = 1$ or $2$. |
| | **M1** | $\frac{\textit{their}\ 5\times4\times3\times2\times1\times5\times4\times3\times2\times1}{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}$ |
| $=\frac{1}{126}$, $0.00794$ | **A1** | Accept $\frac{28800}{362800}$ OE. |
**Total: 5 marks**7
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements are there of the 10 letters in the word REGENERATE?
\item How many different arrangements are there of the 10 letters in the word REGENERATE in which the 4 Es are together and the 2 Rs have exactly 3 letters in between them?
\item Find the probability that a randomly chosen arrangement of the 10 letters in the word REGENERATE is one in which the consonants ( $\mathrm { G } , \mathrm { N } , \mathrm { R } , \mathrm { R } , \mathrm { T }$ ) and vowels ( $\mathrm { A } , \mathrm { E } , \mathrm { E } , \mathrm { E } , \mathrm { E }$ ) alternate, so that no two consonants are next to each other and no two vowels are next to each other. [5]\\
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 2024 Q7 [10]}}