| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Moderate -0.8 This is a straightforward permutations question requiring systematic counting of arrangements with a constraint (odd numbers). Part (i) is routine calculation (3×2×2=12), and part (ii) extends this to include 1-digit, 2-digit, 3-digit, and 4-digit cases using the same logic. No novel insight required, just careful enumeration of cases. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \(^{**}5\) in \(^3P_2\) ways \(= 6\) | M1 | Recognising ends in 5 or 7, can be implied |
| \(^{**}7\) in \(^3P_2 = 6\); Total 12 AG | A1 [3] | Summing ends in 5 + ends in 7; correct answer following legit working |
| OR listing 457, 547, 467, 647, 567, 657, 475, 745, 465, 645, 675, 765 | M1 | Listing at least 5 different numbers ending in 5 |
| M1 | Listing at least 5 different numbers ending in 7 | |
| Total 12 AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 1 digit in 2 ways; 2 digits in \(^*5\) or \(^*7 = {}^3P_1 \times 2 = 6\) | M1 | Consider at least 3 options with different number of digits; if no working must be 3 or 4 from 2, 6, 12, 12 |
| 4 digits in \(^{*}5\) or \(^{*}7 = {}^3P_3 \times 2 = 12\) | A1 | One option correct from 1, 2 or 4 digits |
| Total ways \(= 32\) | A1 [3] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $^{**}5$ in $^3P_2$ ways $= 6$ | M1 | Recognising ends in 5 or 7, can be implied |
| $^{**}7$ in $^3P_2 = 6$; Total 12 AG | A1 [3] | Summing ends in 5 + ends in 7; correct answer following legit working |
| **OR** listing 457, 547, 467, 647, 567, 657, 475, 745, 465, 645, 675, 765 | M1 | Listing at least 5 different numbers ending in 5 |
| | M1 | Listing at least 5 different numbers ending in 7 |
| Total 12 AG | A1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 1 digit in 2 ways; 2 digits in $^*5$ or $^*7 = {}^3P_1 \times 2 = 6$ | M1 | Consider at least 3 options with different number of digits; if no working must be 3 or 4 from 2, 6, 12, 12 |
| 4 digits in $^{***}5$ or $^{***}7 = {}^3P_3 \times 2 = 12$ | A1 | One option correct from 1, 2 or 4 digits |
| Total ways $= 32$ | A1 [3] | |
---3 Numbers are formed using some or all of the digits 4, 5, 6, 7 with no digit being used more than once.\\
(i) Show that, using exactly 3 of the digits, there are 12 different odd numbers that can be formed.\\
(ii) Find how many odd numbers altogether can be formed.
\hfill \mbox{\textit{CAIE S1 2016 Q3 [6]}}