| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Exactly N letters between items |
| Difficulty | Standard +0.3 Part (a)(i) is a standard permutations with repetition problem (10!/(2!2!2!)). Part (a)(ii) requires treating the Ws as fixed positions with 6 letters between them, which is a straightforward extension. Part (b) involves systematic casework with combinations under multiple constraints, but the cases are manageable and the techniques are standard for S1 level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{10!}{2!3!} = 302400\) | B1 [1] | Exact value only, isw rounding |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. \*W\*\*\*\*\*\*W\*, \*\*W\*\*\*\*\*\*W, W\*\*\*\*\*\*W\*\* | M1 | 8! Seen mult or alone. Cannot be embedded (arrangements of other 8 letters) |
| \(\dfrac{8!}{3!} \times 3\) (for the Ws) | M1 | Dividing by \(3!\) (removing repeated L's) |
| M1 | Mult by 3 (different W positions) may be sum of 3 terms | |
| \(= 20160\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| S(5) A(7) C(4) combinations: \(1,3,2: 5 \times {}^7C_3 \times {}^4C_2 = 1050\) | M1 | Mult 3 combinations, \({}^5C_x\), \({}^7C_y\), \({}^4C_z\) (not \(5 \times 7 \times 4\)) |
| \(1,4,1: 5 \times {}^7C_4 \times 4 = 700\) | ||
| \(2,3,1: {}^5C_2 \times {}^7C_3 \times 4 = 1400\) | A1 | 2 correct options unsimplified |
| \(3,2,1: {}^5C_3 \times {}^7C_2 \times 4 = 840\) | M1 | Summing only 3 or 4 correct outcomes involving combs or perms |
| Total \(= 3990\) | A1 [4] |
## Question 7:
### Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{10!}{2!3!} = 302400$ | B1 [1] | Exact value only, isw rounding |
### Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. \*W\*\*\*\*\*\*W\*, \*\*W\*\*\*\*\*\*W, W\*\*\*\*\*\*W\*\* | M1 | 8! Seen mult or alone. Cannot be embedded (arrangements of other 8 letters) |
| $\dfrac{8!}{3!} \times 3$ (for the Ws) | M1 | Dividing by $3!$ (removing repeated L's) |
| | M1 | Mult by 3 (different W positions) may be sum of 3 terms |
| $= 20160$ | A1 [4] | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| S(5) A(7) C(4) combinations: $1,3,2: 5 \times {}^7C_3 \times {}^4C_2 = 1050$ | M1 | Mult 3 combinations, ${}^5C_x$, ${}^7C_y$, ${}^4C_z$ (not $5 \times 7 \times 4$) |
| $1,4,1: 5 \times {}^7C_4 \times 4 = 700$ | | |
| $2,3,1: {}^5C_2 \times {}^7C_3 \times 4 = 1400$ | A1 | 2 correct options unsimplified |
| $3,2,1: {}^5C_3 \times {}^7C_2 \times 4 = 840$ | M1 | Summing only 3 or 4 correct outcomes involving combs or perms |
| Total $= 3990$ | A1 [4] | |7
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements which can be made of all 10 letters of the word WALLFLOWER if
\begin{enumerate}[label=(\roman*)]
\item there are no restrictions,
\item there are exactly six letters between the two Ws.
\end{enumerate}\item A team of 6 people is to be chosen from 5 swimmers, 7 athletes and 4 cyclists. There must be at least 1 from each activity and there must be more athletes than cyclists. Find the number of different ways in which the team can be chosen.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2016 Q7 [9]}}