| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | People arrangements in groups/rows |
| Difficulty | Standard +0.8 This is a multi-part combinatorics problem requiring careful case analysis in part (i) to handle the 'at least' constraints, and non-trivial arrangement techniques in part (iii) involving the complement or gap method for non-adjacent poodles. While the individual techniques are A-level standard, the combination of selection constraints and arrangement restrictions with multiple conditions makes this moderately challenging. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(S(10) \, R(14) \, P(6)\) | M1 | Summing 2 or more 3-factor options perms or combs |
| \(\frac{1}{2} \times \frac{2}{2} \times \frac{4}{1} = 10C1 \times 14C2 \times 6C4 = 13650\) | M1 | Mult 3 combs or 4 combs with \(\Sigma=7\) |
| \(\frac{1}{3} \times \frac{3}{1} = 10C2 \times 14C3 \times 6C3 = 72800\) | 2 options correct, unsimplified | |
| \(\frac{2}{2} \times \frac{3}{1} = 10C2 \times 14C2 \times 6C3 = 81900\) | B1 | |
| Total \(= 168350\) or \(168000\) | A1 [4] | Correct answer |
| (ii) \(2! \times 2! \times 5!\) | M1 | \(2! \times 2!\) oe, seen mult by an integer \(\geq 1\), no division |
| M1 | Mult by \(5!\), or \(5!\) alone, seen mult by an integer \(\geq 1\) no division | |
| \(= 480\) | A1 [3] | Correct answer |
| If M0 earned \(\frac{2! \times 2!}{2! \times 2!}\) or \(\frac{5!}{3!}\) or both, seen mult by an integer \(\geq 1\) | SCM1 | |
| (iii) spaniels and retrievers in 4! ways; gaps in 5P3 or 5 × 4 × 3 ways | M1 | 4! seen multiplied by an integer \(> 1\) |
| \(= 1440\) | M1 | Mult by 5P3 oe |
| A1 [3] | Correct answer | |
| If M0 earned | SCM1 | \(_5C_3\) oe |
| \(\frac{4!}{2!×2!}\) or \(\frac{_5P_3}{3!}\) or both, seen multiplied by an integer \(> 1\) | ||
| or | ||
| \(7! - 5! \times 3! - \{(4! \times 2 \times 4 \times 3!) + (4! \times 3 \times 4 \times 3!)\}\) | M1 | oe |
| \(= 1440\) | M1 | oe, e.g. 6 × 5 × 4 × 4! |
| A1 | ||
| If M0 earned | SCM1 | Marks cannot be earned from both methods. |
| \(3! \times 2! \times 2!\) used as a denominator in all 4 terms |
**(i)** $S(10) \, R(14) \, P(6)$ | M1 | Summing 2 or more 3-factor options perms or combs
$\frac{1}{2} \times \frac{2}{2} \times \frac{4}{1} = 10C1 \times 14C2 \times 6C4 = 13650$ | M1 | Mult 3 combs or 4 combs with $\Sigma=7$
$\frac{1}{3} \times \frac{3}{1} = 10C2 \times 14C3 \times 6C3 = 72800$ | | 2 options correct, unsimplified
$\frac{2}{2} \times \frac{3}{1} = 10C2 \times 14C2 \times 6C3 = 81900$ | B1 |
Total $= 168350$ or $168000$ | A1 [4] | Correct answer
**(ii)** $2! \times 2! \times 5!$ | M1 | $2! \times 2!$ oe, seen mult by an integer $\geq 1$, no division
| M1 | Mult by $5!$, or $5!$ alone, seen mult by an integer $\geq 1$ no division
$= 480$ | A1 [3] | Correct answer
If M0 earned $\frac{2! \times 2!}{2! \times 2!}$ or $\frac{5!}{3!}$ or both, seen mult by an integer $\geq 1$ | SCM1 |
**(iii)** spaniels and retrievers in 4! ways; gaps in 5P3 or 5 × 4 × 3 ways | M1 | 4! seen multiplied by an integer $> 1$
$= 1440$ | M1 | Mult by 5P3 oe
| A1 [3] | Correct answer
If M0 earned | SCM1 | $_5C_3$ oe
$\frac{4!}{2!×2!}$ or $\frac{_5P_3}{3!}$ or both, seen multiplied by an integer $> 1$ | |
or | |
$7! - 5! \times 3! - \{(4! \times 2 \times 4 \times 3!) + (4! \times 3 \times 4 \times 3!)\}$ | M1 | oe
$= 1440$ | M1 | oe, e.g. 6 × 5 × 4 × 4!
| A1 | |
If M0 earned | SCM1 | Marks cannot be earned from both methods.
$3! \times 2! \times 2!$ used as a denominator in all 4 terms | |7 There are 10 spaniels, 14 retrievers and 6 poodles at a dog show. 7 dogs are selected to go through to the final.\\
(i) How many selections of 7 different dogs can be made if there must be at least 1 spaniel, at least 2 retrievers and at least 3 poodles?
2 spaniels, 2 retrievers and 3 poodles go through to the final. They are placed in a line.\\
(ii) How many different arrangements of these 7 dogs are there if the spaniels stand together and the retrievers stand together?\\
(iii) How many different arrangements of these 7 dogs are there if no poodle is next to another poodle?
\hfill \mbox{\textit{CAIE S1 2013 Q7 [10]}}