| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Items NOT together (general separation) |
| Difficulty | Standard +0.3 This is a multi-part combinatorics question covering standard A-level techniques: (a) uses complement counting for non-adjacent arrangements, (b) is straightforward case enumeration with selection, and (c) applies inclusion-exclusion to password constraints. All parts follow textbook methods with no novel insight required, making it slightly easier than average despite being multi-step. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: e.g. xxxxx \(= 5!\) for the other children | (B1 | \(5!\) OE seen alone or multiplied by integer \(k \geqslant 1\), no addition |
| Put y in 6 ways, then 5 then 4 for the youngest children | B1 | Multiply by \(6P3\) OE |
| \(5! \times 6P3 = 14400\) | B1) | Correct answer |
| OR: total \(- 3\) tog \(- 2\) tog \(= 8! - 6!3! - 6! \times 2 \times 5 \times 3 = 14400\) | (B1 | \(8! - 6! \times k \geqslant 1\) seen |
| B1 | \(6!3!\) or \(6! \times 2 \times 5 \times 3\) seen subtracted | |
| B1) | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| D=2, W=2, M=1: \(6C2 \times 4C2 \times 1 = 90\) | B1 | One correct unsimplified option |
| D=3, W=1, M=1: \(6C3 \times 4 \times 1 = 80\) | M1 | Summing 2 or more 3-factor options which can contain terms of 3 factors added; the 1 can be implied |
| D=1, W=3, M=1: \(6 \times 4C3 \times 1 = 24\) | M1 | Summing the correct 3 unsimplified outcomes only |
| Total \(= 194\) ways | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| C=2, D=1, S=1: \(^{26}C_2 \times 9 \times 5 \times 4! = 351000\) | M1 | Summing 2 or more options of the form (2 1 1), (1 2 1), (1 1 2); can have perms, can be added |
| C=1, D=2, S=1: \(26 \times {}^9C_2 \times 5 \times 4! = 112320\) | M1 | 4 relevant products seen excluding 4!; e.g. \(26 \times 9 \times 8 \times 5\) or \(26 \times {}^9P_2 \times 5\) for 2nd outcome; condone \(26 \times 9 \times 5 \times 37\) as being relevant |
| C=1, D=1, S=2: \(26 \times 9 \times {}^5C_2 \times 4! = 56160\) | M1 | Multiply all terms by \(4!\) or \(4!/2!\) |
| Total \(= 519480\) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| EITHER: e.g. xxxxx $= 5!$ for the other children | (B1 | $5!$ OE seen alone or multiplied by integer $k \geqslant 1$, no addition |
| Put y in 6 ways, then 5 then 4 for the youngest children | B1 | Multiply by $6P3$ OE |
| $5! \times 6P3 = 14400$ | B1) | Correct answer |
| OR: total $- 3$ tog $- 2$ tog $= 8! - 6!3! - 6! \times 2 \times 5 \times 3 = 14400$ | (B1 | $8! - 6! \times k \geqslant 1$ seen |
| | B1 | $6!3!$ or $6! \times 2 \times 5 \times 3$ seen subtracted |
| | B1) | Correct answer |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| D=2, W=2, M=1: $6C2 \times 4C2 \times 1 = 90$ | B1 | One correct unsimplified option |
| D=3, W=1, M=1: $6C3 \times 4 \times 1 = 80$ | M1 | Summing 2 or more 3-factor options which can contain terms of 3 factors added; the 1 can be implied |
| D=1, W=3, M=1: $6 \times 4C3 \times 1 = 24$ | M1 | Summing the correct 3 unsimplified outcomes only |
| Total $= 194$ ways | A1 | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| C=2, D=1, S=1: $^{26}C_2 \times 9 \times 5 \times 4! = 351000$ | M1 | Summing 2 or more options of the form (2 1 1), (1 2 1), (1 1 2); can have perms, can be added |
| C=1, D=2, S=1: $26 \times {}^9C_2 \times 5 \times 4! = 112320$ | M1 | 4 relevant products seen excluding 4!; e.g. $26 \times 9 \times 8 \times 5$ or $26 \times {}^9P_2 \times 5$ for 2nd outcome; condone $26 \times 9 \times 5 \times 37$ as being relevant |
| C=1, D=1, S=2: $26 \times 9 \times {}^5C_2 \times 4! = 56160$ | M1 | Multiply all terms by $4!$ or $4!/2!$ |
| Total $= 519480$ | A1 | |
**Total: 4 marks**7
\begin{enumerate}[label=(\alph*)]
\item Eight children of different ages stand in a random order in a line. Find the number of different ways this can be done if none of the three youngest children stand next to each other.
\item David chooses 5 chocolates from 6 different dark chocolates, 4 different white chocolates and 1 milk chocolate. He must choose at least one of each type. Find the number of different selections he can make.
\item A password for Chelsea's computer consists of 4 characters in a particular order. The characters are chosen from the following.
\begin{itemize}
\item The 26 capital letters A to Z
\item The 9 digits 1 to 9
\item The 5 symbols \# \~{} * ? !
\end{itemize}
The password must include at least one capital letter, at least one digit and at least one symbol. No character can be repeated. Find the number of different passwords that Chelsea can make.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2017 Q7 [11]}}