| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with identical objects |
| Difficulty | Moderate -0.8 This is a straightforward permutations question testing standard formulas for arrangements with repetition (multinomial coefficients) and basic combinations. Part (i) is direct application of n!/(n₁!n₂!...nₖ!), part (ii) splits the problem into two independent groups using the same formula, and part (iii) is simple combination multiplication. All parts require only routine recall and calculation with no problem-solving insight or novel reasoning. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Road | |||
| \(\square \square \square \square \square \square \square \square \square\) | \(\square \square \square\) | ||
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{12!}{2!2!3!4!} = 831600\) | M1 A1 [2] | Dividing by 3! 4! and 2! once or twice o.e; correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{6!}{4!2!} \times \frac{6!}{2!3!}\) | B1 | \(\frac{6!}{4!2!}\) and \(\frac{6!}{2!3!}\) seen o.e |
| M1 | Multiplying their numbers for group 1 with their numbers for group 2 | |
| \(= 900\) | A1 [3] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2 \times 3 \times {}_{7}C_2\) or \(2 \times 3 \times 21\) | M1 | \({}_{7}C_2\) seen multiplied or 5 options added |
| \(= 126\) | A1 [2] | Correct final answer |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{12!}{2!2!3!4!} = 831600$ | M1 A1 **[2]** | Dividing by 3! 4! and 2! once or twice o.e; correct final answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{6!}{4!2!} \times \frac{6!}{2!3!}$ | B1 | $\frac{6!}{4!2!}$ and $\frac{6!}{2!3!}$ seen o.e |
| | M1 | Multiplying their numbers for group 1 with their numbers for group 2 |
| $= 900$ | A1 **[3]** | Correct final answer |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2 \times 3 \times {}_{7}C_2$ or $2 \times 3 \times 21$ | M1 | ${}_{7}C_2$ seen multiplied or 5 options added |
| $= 126$ | A1 **[2]** | Correct final answer |
---4 A builder is planning to build 12 houses along one side of a road. He will build 2 houses in style $A$, 2 houses in style $B , 3$ houses in style $C , 4$ houses in style $D$ and 1 house in style $E$.\\
(i) Find the number of possible arrangements of these 12 houses.\\
(ii)
\begin{center}
\begin{tabular}{ c c c }
\hline
\multicolumn{4}{c}{Road} \\
\hline
$\square \square \square \square \square \square \square \square \square$ & $\square \square \square$ & & \\
\hline
\end{tabular}
\end{center}
The 12 houses will be in two groups of 6 (see diagram). Find the number of possible arrangements if all the houses in styles $A$ and $D$ are in the first group and all the houses in styles $B , C$ and $E$ are in the second group.\\
(iii) Four of the 12 houses will be selected for a survey. Exactly one house must be in style $B$ and exactly one house in style $C$. Find the number of ways in which these four houses can be selected.
\hfill \mbox{\textit{CAIE S1 2008 Q4 [7]}}