| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Specific items together |
| Difficulty | Moderate -0.3 This is a standard permutations question with two classic scenarios (grouping items together, separating items). Part (a) requires treating 4 men as a single unit (7! × 4!), part (b) uses the gaps method (6! × ⁷P₄). Both are textbook techniques with straightforward application, making it slightly easier than average but still requiring proper combinatorial reasoning. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(7! \times 4!\) | M1 | Allow for \(6!\times 4!\) or \(6!\times 4!\times 2\) |
| \(\div\ 10!\) | M1 | Divide by \(10!\), needs at least one factorial in numerator |
| \(= \dfrac{120960}{3628800} = \dfrac{1}{30}\) | A1 [3] | Answer, exact or awrt 0.0333 |
| Answer | Marks | Guidance |
|---|---|---|
| \(7\times\dfrac{6}{10}\times\dfrac{4}{9}\times\dfrac{5}{8}\times\dfrac{3}{7}\times\dfrac{4}{6}\times\dfrac{2}{5}\times\dfrac{3}{4}\times\dfrac{1}{3}\times\dfrac{2}{2}\times\dfrac{1}{1}\) | M1 M1 A1 [3] | no 7, one other error; only one error; correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Women placed in \(6!\) ways, men in \(4!\) \([= 720\times 24]\) | B1 | \(6!\times 4!\) anywhere, or \(6!\times\text{attempt at } {}^7P_4\) |
| 4 slots \(m\) in \(mWmWmWmWmWm = {}^7C_4\) | M1 | Or \({}^7P_4\). Allow for \(m\) and \(W\) reversed |
| \({}^7C_4\times\dfrac{6!\times 4!}{10!}\) | M1 | Needs attempt at both terms |
| \(= \dfrac{1}{6}\) | A1 [3] | Or 0.167 or 0.1667 etc |
| Answer | Marks | Guidance |
|---|---|---|
| \((10! - 12\times9! + (3\times4\times8! + 12\times2\times8!) - 24\times7!)/10!\) | M2 | Signs alternating, at least one term \(\checkmark\) |
| \([=(3628800-4354560+1451520-120960)/10!]\) | A1 A1 | Allow one term omitted or wrong; Correct answer |
| Three together: \(7\times 6\times\dfrac{6!4!}{10!}=\dfrac{1}{5}\) | Two pairs: \(\dfrac{7\times 6}{2}\times\dfrac{6!4!}{10!}=\dfrac{1}{10}\) | One pair: \(7\times\dfrac{6\times 5}{2}\times\dfrac{6!4!}{10!}=\dfrac{1}{2}\) |
# Question 4:
## Part (a)
$7! \times 4!$ | **M1** | Allow for $6!\times 4!$ or $6!\times 4!\times 2$
$\div\ 10!$ | **M1** | Divide by $10!$, needs at least one factorial in numerator
$= \dfrac{120960}{3628800} = \dfrac{1}{30}$ | **A1 [3]** | Answer, exact or awrt 0.0333 | 3/3 for $\dfrac{1}{30}$ www
**Alternative:**
$7\times\dfrac{6}{10}\times\dfrac{4}{9}\times\dfrac{5}{8}\times\dfrac{3}{7}\times\dfrac{4}{6}\times\dfrac{2}{5}\times\dfrac{3}{4}\times\dfrac{1}{3}\times\dfrac{2}{2}\times\dfrac{1}{1}$ | **M1 M1 A1 [3]** | no 7, one other error; only one error; correct answer
## Part (b)
Women placed in $6!$ ways, men in $4!$ $[= 720\times 24]$ | **B1** | $6!\times 4!$ anywhere, or $6!\times\text{attempt at } {}^7P_4$ | Or $6!\times 7\times 6\times 5\times 4$ B2
4 slots $m$ in $mWmWmWmWmWm = {}^7C_4$ | **M1** | Or ${}^7P_4$. Allow for $m$ and $W$ reversed | ${}^7P_4\times 6!$: B1M1
${}^7C_4\times\dfrac{6!\times 4!}{10!}$ | **M1** | Needs attempt at both terms | $4\times(6!\times 4!)/10! = 2/105$: B1M1
$= \dfrac{1}{6}$ | **A1 [3]** | Or 0.167 or 0.1667 etc
**Alternative PIE:**
$(10! - 12\times9! + (3\times4\times8! + 12\times2\times8!) - 24\times7!)/10!$ | **M2** | Signs alternating, at least one term $\checkmark$
$[=(3628800-4354560+1451520-120960)/10!]$ | **A1 A1** | Allow one term omitted or wrong; Correct answer
Three together: $7\times 6\times\dfrac{6!4!}{10!}=\dfrac{1}{5}$ | Two pairs: $\dfrac{7\times 6}{2}\times\dfrac{6!4!}{10!}=\dfrac{1}{10}$ | One pair: $7\times\dfrac{6\times 5}{2}\times\dfrac{6!4!}{10!}=\dfrac{1}{2}$
---4 The members of a team stand in a random order in a straight line for a photograph. There are four men and six women.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that all the men are next to each other.
\item Find the probability that no two men are next to one another.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2019 Q4 [7]}}