| 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 +0.3 This is a standard permutations question with repeated letters (SEVENTEEN has 4 Es, 2 Ns). Parts (i)-(ii) involve straightforward arrangements with constraints using factorial division for repeated elements. Parts (iii)-(iv) are basic combinations with simple case-work. All techniques are routine for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| E with other letters arranged in \(\frac{8!}{2!3!}\) | M1 | Multiply by \(8!\) or \(^8P_8\) oe (arrangements ignoring repeats) |
| \(= 3360\) ways | A1 | Correct final answer www |
| OR: \(\frac{8\times7\times6\times5\times4\times4\times3\times2\times1}{4!2!} = 3360\) ways | M1 | Correct numerator (161 280) |
| A1 | Correct final answer www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| * * * * * with arrangements of other letters \(\times\) ways Es inserted | M1 | \(k\) mult by \(^6C_4\) or \(^6P_4\) oe (ways to insert Es ignoring repeats), \(k\) can \(= 1\), or \(k\) mult by \(\frac{5!}{2!}\) |
| \(= \frac{5!}{2!} \times {^6C_4}\left(\frac{5!}{2!}\times\frac{^6P_4}{4!}\right)\) | M1 | Correct unsimplified expression or \(\frac{5!}{2!} \times {^6P_4}\) |
| \(= 900\) ways | A1 | Correct answer |
| OR: Total \(-\) no of ways with Es touching: \(9!/(4!\times2!) - \ldots\) or \(7560 - \ldots\) | M1 | \(7560\) unsimplified \(- k\) |
| \(\frac{6!}{2!} + {^6P_2}\times\frac{5!}{2!} + \frac{^6P_2}{2!}\times\frac{5!}{2!} + \frac{^6P_3}{2!\times\frac{5!}{2!}}\) \(= 360 + 1800 + 900 + 3600 = 6660\) | M1 | Attempting to find four ways of Es touching (4 Es, 3Es and a single, 2 lots of 2 Es, 2 Es and 2 singles) |
| \(7560 - 6660 = 900\) | A1 | Correct answer |
| OR: \(\frac{5!}{2!}(E_1 + E_2 + E_3)\) where \(E_1=10, E_2=4, E_3=1\) | M1 | For any values for \(E_1\), \(E_2\) and \(E_3\) |
| \(\frac{5!}{2!}(E_1+E_2+E_3)\) | M1 | For any two correct values of \(E_1\), \(E_2\) and \(E_3\) |
| \(600 + 240 + 60 = 900\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| EENN* in 3 ways | B1 | Numerical value must be stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| EE* with no N: 1 way; EEN 3C2 or listing 3 ways; EENN* 3 ways from (iii) | M1 | Identifying the three different scenarios of EE, EEE or EEEE |
| A1 | Total no of ways with two Es (\(7\) or \(3+3+1\)) | |
| EEE** with no N: 3 ways; EEEN* 3 ways; EEENN 1 way | A1 | Total no. of ways with 3 Es (7) |
| EEEE* no N 3 ways; EEEEN 1 way; Total 18 ways | A1 | Correct answer stated |
| Method: List containing ways with 2Es, 3Es and 4Es; List containing at least 8 correct different ways; List of all 18 correct ways; Total 18 | M1 | At least 1 option listed for each of EE\*\*\*, EEE\*\*, EEEE\* |
| A1 | Ignore repeated options | |
| A1 | Ignore repeated/incorrect options | |
| A1 | Correct answer stated |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| ****E**** with other letters arranged in $\frac{8!}{2!3!}$ | M1 | Multiply by $8!$ or $^8P_8$ oe (arrangements ignoring repeats) |
| $= 3360$ ways | A1 | Correct final answer www |
| OR: $\frac{8\times7\times6\times5\times4\times4\times3\times2\times1}{4!2!} = 3360$ ways | M1 | Correct numerator (161 280) |
| | A1 | Correct final answer www |
**Total: 2**
---
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| * * * * * with arrangements of other letters $\times$ ways Es inserted | M1 | $k$ mult by $^6C_4$ or $^6P_4$ oe (ways to insert Es ignoring repeats), $k$ can $= 1$, or $k$ mult by $\frac{5!}{2!}$ |
| $= \frac{5!}{2!} \times {^6C_4}\left(\frac{5!}{2!}\times\frac{^6P_4}{4!}\right)$ | M1 | Correct unsimplified expression or $\frac{5!}{2!} \times {^6P_4}$ |
| $= 900$ ways | A1 | Correct answer |
| OR: Total $-$ no of ways with Es touching: $9!/(4!\times2!) - \ldots$ or $7560 - \ldots$ | M1 | $7560$ unsimplified $- k$ |
| $\frac{6!}{2!} + {^6P_2}\times\frac{5!}{2!} + \frac{^6P_2}{2!}\times\frac{5!}{2!} + \frac{^6P_3}{2!\times\frac{5!}{2!}}$ $= 360 + 1800 + 900 + 3600 = 6660$ | M1 | Attempting to find four ways of Es touching (4 Es, 3Es and a single, 2 lots of 2 Es, 2 Es and 2 singles) |
| $7560 - 6660 = 900$ | A1 | Correct answer |
| OR: $\frac{5!}{2!}(E_1 + E_2 + E_3)$ where $E_1=10, E_2=4, E_3=1$ | M1 | For any values for $E_1$, $E_2$ and $E_3$ |
| $\frac{5!}{2!}(E_1+E_2+E_3)$ | M1 | For any two correct values of $E_1$, $E_2$ and $E_3$ |
| $600 + 240 + 60 = 900$ | A1 | Correct answer |
**Total: 3**
---
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| EENN* in 3 ways | B1 | Numerical value must be stated |
**Total: 1**
---
## Question 7(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| EE*** with no N: 1 way; EEN** 3C2 or listing 3 ways; EENN* 3 ways from **(iii)** | M1 | Identifying the three different scenarios of EE, EEE or EEEE |
| | A1 | Total no of ways with two Es ($7$ or $3+3+1$) |
| EEE** with no N: 3 ways; EEEN* 3 ways; EEENN 1 way | A1 | Total no. of ways with 3 Es (7) |
| EEEE* no N 3 ways; EEEEN 1 way; Total 18 ways | A1 | Correct answer stated |
| **Method:** List containing ways with 2Es, 3Es and 4Es; List containing at least 8 correct different ways; List of all 18 correct ways; Total 18 | M1 | At least 1 option listed for each of EE\*\*\*, EEE\*\*, EEEE\* |
| | A1 | Ignore repeated options |
| | A1 | Ignore repeated/incorrect options |
| | A1 | Correct answer stated |
**Total: 4**7 Find the number of ways the 9 letters of the word SEVENTEEN can be arranged in each of the following cases.\\
(i) One of the letter Es is in the centre with 4 letters on either side.\\
(ii) No E is next to another E.\\
5 letters are chosen from the 9 letters of the word SEVENTEEN.\\
(iii) Find the number of possible selections which contain exactly 2 Es and exactly 2 Ns.\\
(iv) Find the number of possible selections which contain at least 2 Es.\\
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]}}