| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with alternating patterns |
| Difficulty | Standard +0.8 Part (i) is routine permutations with repetition. Part (ii) requires systematic case-work to ensure vowels alternate with consonants, which is non-trivial. Parts (iii-iv) involve selection with restrictions requiring careful counting of cases with repeated letters. The combination of arrangement constraints and selection problems with repetitions makes this moderately challenging, requiring more problem-solving than standard textbook exercises. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{9!}{2!2!} = 90720\) | B1 | Must see 90720 |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: arrangement shown with consonants and vowel slots | B1 | \(5!\) seen multiplied (arrangement of consonants allowing repeats) |
| No. arrangements of consonants \(\times\) ways of inserting vowels | B1 | \(^6P_4\) oe (i.e. \(6\times5\times4\times3\), \(^6C_4\times4!\)) seen multiplied (allowing repeats), no extra terms |
| \(\frac{5!}{2!} \times \frac{^6P_4}{2!}\) | B1 | Dividing by at least one \(2!\) (removing at least one set of repeats) |
| \(\frac{^6P_4}{2!}\times\frac{5}{2} = 10\,800\) | B1 | Correct final answer |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^5C_3 = 10\) | M1 | \(^5C_x\) or \(^5P_x\) seen alone, \(x=2\) or \(3\) |
| A1 | Correct final answer not from \(^5C_2\) | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: Considering separate groups | M1 | Considering two scenarios of MME or EEM or MMEE with attempt, may be probabilities or permutations |
| \(\text{MME} = {^5C_2} = 10\); \(\text{MEE} = {^5C_2} = 10\); \(\text{MMEE*} = {^5C_1} = 5\) | M1 | Summing three appropriate scenarios from the four; need \(^5C_x\) seen in all of them |
| \(\text{ME*} = {^5C_3} = 10\) see (iii); Total \(= 35\) | A1** | Correct final answer |
| Method 2: Considering criteria met if ME are chosen | M1 | \(^7C_x\) only seen, no other terms |
| M1 | \(^7C_3\) only seen, no other terms | |
| \(\text{ME*} = {^7C_3} = 35\) | A1** | Correct final answer |
| [3] |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{9!}{2!2!} = 90720$ | **B1** | Must see 90720 |
| | **[1]** | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Method 1: arrangement shown with consonants and vowel slots | **B1** | $5!$ seen multiplied (arrangement of consonants allowing repeats) |
| No. arrangements of consonants $\times$ ways of inserting vowels | **B1** | $^6P_4$ oe (i.e. $6\times5\times4\times3$, $^6C_4\times4!$) seen multiplied (allowing repeats), no extra terms |
| $\frac{5!}{2!} \times \frac{^6P_4}{2!}$ | **B1** | Dividing by at least one $2!$ (removing at least one set of repeats) |
| $\frac{^6P_4}{2!}\times\frac{5}{2} = 10\,800$ | **B1** | Correct final answer |
| | **[4]** | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^5C_3 = 10$ | **M1** | $^5C_x$ or $^5P_x$ seen alone, $x=2$ or $3$ |
| | **A1** | Correct final answer not from $^5C_2$ |
| | **[2]** | |
---
## Question 7(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** Considering separate groups | **M1** | Considering two scenarios of MME or EEM or MMEE with attempt, may be probabilities or permutations |
| $\text{MME**} = {^5C_2} = 10$; $\text{MEE**} = {^5C_2} = 10$; $\text{MMEE*} = {^5C_1} = 5$ | **M1** | Summing three appropriate scenarios from the four; need $^5C_x$ seen in all of them |
| $\text{ME***} = {^5C_3} = 10$ see (iii); Total $= 35$ | **A1** | Correct final answer |
| **Method 2:** Considering criteria met if ME are chosen | **M1** | $^7C_x$ only seen, no other terms |
| | **M1** | $^7C_3$ only seen, no other terms |
| $\text{ME***} = {^7C_3} = 35$ | **A1** | Correct final answer |
| | **[3]** | |7 Find the number of different ways in which all 9 letters of the word MINCEMEAT can be arranged in each of the following cases.\\
(i) There are no restrictions.\\
(ii) No vowel (A, E, I are vowels) is next to another vowel.\\
5 of the 9 letters of the word MINCEMEAT are selected.\\
(iii) Find the number of possible selections which contain exactly 1 M and exactly 1 E .\\
(iv) Find the number of possible selections which contain at least 1 M and at least 1 E .\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S1 2018 Q7 [10]}}