## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{9!}{2!2!} =\right]\ 90720$ | B1 | |

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## Question 6(b):

**Method 1**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{9!}{2!2!} - \frac{7!}{2!} \times 2$ | M1 | $a - \frac{7!}{2!} \times b$, $a =$ their 6(a) or correct, $b = 1, 2$ |
| | M1 | $a - \frac{7!}{c!} \times 2$, $a =$ their 6(a) or correct, $c = 1, 2$ |
| $85680$ | A1 FT | ft their $\mathbf{6(a)} - 5040$ |

**Method 2**

| Answer | Marks | Guidance |
|--------|-------|----------|
| P and S at ends: $2 \times 7! = 10080$ | M1 | Finding correct number of ways for one of these correctly identified scenarios |
| P or S at one end only: $4 \times 5 \times \frac{7!}{2!} = 50400$ | M1 | Adding no of ways for 3 correctly identified scenarios |
| Neither P nor S at an end: $5 \times 4 \times \frac{7!}{2!2!} = 25200$ | | |
| Total: $85680$ | A1 | |

**Method 3**

| Answer | Marks | Guidance |
|--------|-------|----------|
| P at beginning: $7 \times \frac{7!}{2!} = 17640$ | M1 | Finding correct number of ways for one of these correctly identified scenarios |
| S at beginning: $7 \times \frac{7!}{2!} = 17640$ | M1 | Adding no of ways for 3 correctly identified scenarios |
| Neither P nor S at beginning: $5 \times \frac{8!}{2!2!} = 50400$ | | |
| Total: $85680$ | A1 | |

## Question 6(c):

**Method 1** – arrangements with PP between Ss

| Answer/Working | Marks | Guidance |
|---|---|---|
| $6! + 5! \times 5 \times 4$ | M1 | $6! + d$, $d$ an integer $\geq 1$, may be implied |
| $6! + 5! \times 5 \times 4$ | M1 | $e + 5! \times f$, $e, f$ integers $\geq 1$, may be implied |
| | M1 | $e + g! \times (5 \times 4 \text{ or } {}^5P_2)$, $e$ an integer $\geq 1$, $g = 4, 5, 6$ |
| $[\text{Total}] = 3120$ | A1 | |

**Method 2** – considers the 6 positions for S^^S

| Answer/Working | Marks | Guidance |
|---|---|---|
| Positions 1 and 6: there are $5 \times 5!$ ways | M1 | Identifying no. of ways if S^^S is in position 1 or 6 |
| Positions 2, 3, 4 and 5: there are $4 \times 5!$ ways | M1 | Identifying no. of ways if S^^S is in position 2, 3, 4 or 5 |
| $2 \times 5 \times 5! + 4 \times 4 \times 5!$ | M1 | Adding no. of ways for 6 scenarios (or $26 \times 5!$) |
| $[\text{Total}] = 3120$ | A1 | SC B1 for 3120 if any method marks are withheld |

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## Question 6(d):

**Method 1** – Either PP in group of 5 or PP in group of 4

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{{}^5C_3}{{}^9C_5} + \dfrac{{}^5C_2}{{}^9C_5}$ | M1 | $a \times {}^5C_2$, $a \times {}^5C_3$, or ${}^5C_2 + {}^5C_3$ seen as numerator of one or two fractions, where $a$ is 1 or 2, no extra terms |
| $\dfrac{{}^5C_3 + {}^5C_2}{{}^9C_5}$ | M1 | ${}^9C_5$ or ${}^9C_4$ seen as a denominator of one or two fractions |
| Probability $= \dfrac{20}{126}, \dfrac{10}{63}, 0.159$ | A1 | |

**Method 2** – Considering the positions of P and then S

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7} \times \dfrac{3}{6}\right) + \left(\dfrac{5}{7} \times \dfrac{4}{6} \times \dfrac{4}{9} \times \dfrac{3}{8}\right)$ | M1 | $a \times 5 \times 4 \times 4 \times 3$ seen as numerator of a fraction, where $a = 1$ or $2$ |
| $\left(\dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7} \times \dfrac{3}{6}\right) + \left(\dfrac{5}{7} \times \dfrac{4}{6} \times \dfrac{4}{9} \times \dfrac{3}{8}\right)$ | M1 | $9 \times 8 \times 7 \times 6$ seen as denominator of a fraction |
| $= \dfrac{10}{63}$ | A1 | |