| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Standard +0.3 This is a standard permutations question with three parts of increasing complexity. Part (i) requires treating odd digits as a block (standard technique), part (ii) needs casework on first and last digits, and part (iii) combines divisibility rules with place value constraints. All parts use routine counting principles taught in S1 with no novel insight required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(5! \times 3!\) or \(6!\) | B1 | \(5!\) or \(3!\) or \(6!\) or seen multiplied or alone |
| \(= 720\) | B1 [2] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(3^{}4,\ 3^{}8,\ 4^{**}8\) | M1 | Considering at least 2 types of 4-figure options ending with 4 or 8 and starting with 3 or 4 |
| \(= 5 \times 4 + 5 \times 4 + 5 \times 4 = 60\) | B1, A1 [3] | One option correct unsimplified can be implied; Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(5,\ {*}5,\ {**}5\) | M1 | Appreciating that the number must end in 5 (can be implied) |
| \(= 1 + 7 + 7^2\) | M1 | Summing numbers ending in 5 with at least 2 different numbers of digits |
| \(= 57\) | A1 [3] | Correct final answer |
## Question 5:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $5! \times 3!$ or $6!$ | B1 | $5!$ or $3!$ or $6!$ or seen multiplied or alone |
| $= 720$ | B1 **[2]** | Correct final answer |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $3^{**}4,\ 3^{**}8,\ 4^{**}8$ | M1 | Considering at least 2 types of 4-figure options ending with 4 or 8 and starting with 3 or 4 |
| $= 5 \times 4 + 5 \times 4 + 5 \times 4 = 60$ | B1, A1 **[3]** | One option correct unsimplified can be implied; Correct final answer |
### Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $5,\ {*}5,\ {**}5$ | M1 | Appreciating that the number must end in 5 (can be implied) |
| $= 1 + 7 + 7^2$ | M1 | Summing numbers ending in 5 with at least 2 different numbers of digits |
| $= 57$ | A1 **[3]** | Correct final answer |
---5 Find how many different numbers can be made from some or all of the digits of the number 1345789 if\\
(i) all seven digits are used, the odd digits are all together and no digits are repeated,\\
(ii) the numbers made are even numbers between 3000 and 5000, and no digits are repeated,\\
(iii) the numbers made are multiples of 5 which are less than 1000 , and digits can be repeated.
\hfill \mbox{\textit{CAIE S1 2014 Q5 [8]}}