| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with adjacency requirements |
| Difficulty | Moderate -0.8 This is a straightforward permutations question testing standard techniques: treating grouped letters as single units, fixing positions, and combinations with restrictions. Part (a) requires basic counting of repeated letters (4E, 2N, 2S) with simple constraints, while part (b) is a routine selection problem. These are textbook exercises requiring only direct application of formulas with no problem-solving insight or multi-step reasoning beyond the immediate setup. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. \*\*(EEEE)\*\*\* | M1 | Multiply by \(6!\) oe |
| Number of ways \(= \dfrac{6!}{2!2!} = 180\) | M1 | Dividing by \(2!2!\) oe |
| A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| S\*\*\*\*\*\*\*T or T\*\*\*\*\*\*\*\*S | M1 | Multiply by \(7!\) or dividing by one of \(2!\) or \(4!\) |
| Number of ways \(= \dfrac{7!}{4!2!} \times 2 = 210\) | M1 | Multiply by 2 |
| A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Exactly one E in \({}^6C_3\) ways | M1 | \({}^6C_x\) as a single answer |
| \(= 20\) | M1 | \({}^xC_3\) as a single answer |
| A1 [3] | Correct answer |
## Question 5 (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. \*\*(EEEE)\*\*\* | M1 | Multiply by $6!$ oe |
| Number of ways $= \dfrac{6!}{2!2!} = 180$ | M1 | Dividing by $2!2!$ oe |
| | A1 [3] | Correct answer |
## Question 5 (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| S\*\*\*\*\*\*\*T or T\*\*\*\*\*\*\*\*S | M1 | Multiply by $7!$ or dividing by one of $2!$ or $4!$ |
| Number of ways $= \dfrac{7!}{4!2!} \times 2 = 210$ | M1 | Multiply by 2 |
| | A1 [3] | Correct answer |
## Question 5 (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Exactly one E in ${}^6C_3$ ways | M1 | ${}^6C_x$ as a single answer |
| $= 20$ | M1 | ${}^xC_3$ as a single answer |
| | A1 [3] | Correct answer |
---5
\begin{enumerate}[label=(\alph*)]
\item Find the number of ways in which all nine letters of the word TENNESSEE can be arranged
\begin{enumerate}[label=(\roman*)]
\item if all the letters E are together,
\item if the T is at one end and there is an S at the other end.
\end{enumerate}\item Four letters are selected from the nine letters of the word VENEZUELA. Find the number of possible selections which contain exactly one E .
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2015 Q5 [9]}}