| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with alternating patterns |
| Difficulty | Standard +0.8 Part (a) requires systematic handling of alternating vowel/consonant arrangements with repeated letters (3 P's, 2 E's), demanding careful case analysis and application of permutation formulas. Part (b)(ii) adds a conditional constraint requiring complementary counting. While individual techniques are standard A-level, the combination of constraints and multi-step reasoning elevates this above routine exercises. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks |
|---|---|
| M1 | Multiply by \(5!\) in numerator |
| Answer | Marks |
|---|---|
| M1 | Divide by \(3!\) or \(2!\) |
| M1 | Multiply by \(6P_4\) or equivalent |
| A1 | [4] |
| Answer | Marks |
|---|---|
| M1 | Multiply 4 combinations of which three are correct |
| A1 | [2] |
| Answer | Marks |
|---|---|
| M1 | Evaluating both in team and subtracting from (i) |
| Answer | Marks |
|---|---|
| M1 | Follow-through their 420, their 240 |
| A1 | [3] |
| Answer | Marks |
|---|---|
| M1 | Summing 2 or 3 options not both in team |
| A1 | 2 or 3 options correct, unsimplified |
| A1 | Correct answer from correct working |
| Answer | Marks |
|---|---|
| M1 | As above, or bowl in bat out + bowl out |
| A1 A1 | [3] |
**(a)**
e.g. P*N*P*P*L
M1 | Multiply by $5!$ in numerator
$\frac{5!}{3! \cdot 2!} \times 6P_4 = 3600$
M1 | Divide by $3!$ or $2!$
M1 | Multiply by $6P_4$ or equivalent
A1 | [4]
**(b) (i)**
$\binom{7}{5} \times \binom{5}{4} \times \binom{2}{1} \times \binom{2}{1} = 420$
M1 | Multiply 4 combinations of which three are correct
A1 | [2]
**(ii)**
Both in team: $\binom{6}{4} \times \binom{4}{3} \times 2 \times 2 = 240$
M1 | Evaluating both in team and subtracting from (i)
$420 - 240 = 180$ ways
M1 | Follow-through their 420, their 240
A1 | [3]
**OR**
Bat in bowl out + bowl in bat out + both out
$= \binom{6}{4} \times \binom{4}{3} \times 2 \times 2 + \binom{6}{5} \times \binom{4}{3} \times 2 \times 2 + \binom{6}{5} \times \binom{4}{4} \times 2 \times 2$
$= 60 + 96 + 24 = 180$ ways
M1 | Summing 2 or 3 options not both in team
A1 | 2 or 3 options correct, unsimplified
A1 | Correct answer from correct working
**OR**
Bat in bowl out + bat out
$= 60 + \binom{6}{5} \times \binom{5}{4} \times 2 \times 2 = 60 + 120 = 180$ ways
M1 | As above, or bowl in bat out + bowl out
A1 A1 | [3]
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5
\begin{enumerate}[label=(\alph*)]
\item Find the number of different ways of arranging all nine letters of the word PINEAPPLE if no vowel (A, E, I) is next to another vowel.
\item A certain country has a cricket squad of 16 people, consisting of 7 batsmen, 5 bowlers, 2 allrounders and 2 wicket-keepers. The manager chooses a team of 11 players consisting of 5 batsmen, 4 bowlers, 1 all-rounder and 1 wicket-keeper.
\begin{enumerate}[label=(\roman*)]
\item Find the number of different teams the manager can choose.
\item Find the number of different teams the manager can choose if one particular batsman refuses to be in the team when one particular bowler is in the team.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2016 Q5 [9]}}