| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question testing standard techniques: arrangements with repeated letters, positional constraints, and basic selection problems. All parts use direct formula application (9!/4!2! for repeated letters, treating grouped letters as one unit, basic combination counting) with no novel problem-solving required. Easier than average A-level due to being purely procedural. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 7560 ways | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| RxxxxxxxG in \(\dfrac{7!}{4!}\) | B1 | \(7!\) alone seen in numerator or \(4!\) alone in denominator. \(\dfrac{7! \times 2}{4! \times 2}\) gets full marks |
| \(= 210\) ways | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. EEEExxxxx in \(\dfrac{6!}{2!}\) | B1 | \(6!\) or \(5! \times 6\) seen in numerator or on own. Can be \(6! \times k\) but not \(6! \pm k\) |
| \(= 360\) ways | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1 R e.g. RVG or RVN or RGN \(= 3\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| no Rs e.g. VGN or 3C3 ways \(= 1\); 2 Rs e.g. RRV or 3C1 ways \(= 3\) | M1 | Summing at least 2 options for R |
| Total \(= 7\) | A1, A1 [3] | Correct outcome for no Rs or 2 Rs evaluated |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 7560 ways | **B1** [1] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| RxxxxxxxG in $\dfrac{7!}{4!}$ | **B1** | $7!$ alone seen in numerator or $4!$ alone in denominator. $\dfrac{7! \times 2}{4! \times 2}$ gets full marks |
| $= 210$ ways | **B1** [2] | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. EEEExxxxx in $\dfrac{6!}{2!}$ | **B1** | $6!$ or $5! \times 6$ seen in numerator or on own. Can be $6! \times k$ but not $6! \pm k$ |
| $= 360$ ways | **B1** [2] | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1 R e.g. RVG or RVN or RGN $= 3$ | **B1** [1] | |
### Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| no Rs e.g. VGN or 3C3 ways $= 1$; 2 Rs e.g. RRV or 3C1 ways $= 3$ | **M1** | Summing at least 2 options for R |
| Total $= 7$ | **A1**, **A1** [3] | Correct outcome for no Rs or 2 Rs evaluated |
---6 Find the number of ways all 9 letters of the word EVERGREEN can be arranged if\\
(i) there are no restrictions,\\
(ii) the first letter is R and the last letter is G ,\\
(iii) the Es are all together.
Three letters from the 9 letters of the word EVERGREEN are selected.\\
(iv) Find the number of selections which contain no Es and exactly 1 R .\\
(v) Find the number of selections which contain no Es.
\hfill \mbox{\textit{CAIE S1 2016 Q6 [9]}}