| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Moderate -0.3 This is a straightforward permutations question with standard counting techniques. Part (a) requires basic multiplication principle with distinct digits (no complex casework), part (b) is a simple stars-and-bars problem with constraints that reduce to counting 2+2+1, 2+1+2, or 1+2+2 distributions then selecting from available flowers. All techniques are routine for A-level statistics students with no novel problem-solving required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(9\times9\times8 = 648\) | M1 M1, A1 [3] | Logical listing attempt |
| OR \(900 - 28\times9 = 648\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(7\ldots\text{in}\ 1\times8\times4 = 32\) ways | M1 | Listing numbers starting with 7 or 9 and ending odd |
| \(8\ldots\text{in}\ 1\times8\times5 = 40\) | M1 | |
| \(9\ldots\text{in}\ 1\times8\times4 = 32\) | M1 | |
| Total 104 ways | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R(6)\ T(5)\ D(4)\): | ||
| \(2\ 2\ 3 = {}^6C_2\times{}^5C_2\times{}^4C_3 = 600\) | M1 | Mult 3 combs, \({}^6C_x\times{}^5C_y\times{}^4C_z\); Summing 2 or 3 three-factor outcomes |
| \(2\ 3\ 2 = {}^6C_2\times{}^5C_3\times{}^4C_2 = 900\) | M1 | can be perms, \(+\) instead of \(\times\) |
| \(3\ 2\ 2 = {}^6C_3\times{}^5C_2\times{}^4C_2 = 1200\) | A1 | 2 options correct unsimplified |
| Total \(= 2700\) | A1 [4] |
## Question 6:
### Part (a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $9\times9\times8 = 648$ | M1 M1, A1 [3] | Logical listing attempt |
| OR $900 - 28\times9 = 648$ | | |
### Part (a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $7\ldots\text{in}\ 1\times8\times4 = 32$ ways | M1 | Listing numbers starting with 7 or 9 and ending odd |
| $8\ldots\text{in}\ 1\times8\times5 = 40$ | M1 | |
| $9\ldots\text{in}\ 1\times8\times4 = 32$ | M1 | |
| Total 104 ways | A1 [4] | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R(6)\ T(5)\ D(4)$: | | |
| $2\ 2\ 3 = {}^6C_2\times{}^5C_2\times{}^4C_3 = 600$ | M1 | Mult 3 combs, ${}^6C_x\times{}^5C_y\times{}^4C_z$; Summing 2 or 3 three-factor outcomes |
| $2\ 3\ 2 = {}^6C_2\times{}^5C_3\times{}^4C_2 = 900$ | M1 | can be perms, $+$ instead of $\times$ |
| $3\ 2\ 2 = {}^6C_3\times{}^5C_2\times{}^4C_2 = 1200$ | A1 | 2 options correct unsimplified |
| Total $= 2700$ | A1 [4] | |
---6
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find how many numbers there are between 100 and 999 in which all three digits are different.
\item Find how many of the numbers in part (i) are odd numbers greater than 700 .
\end{enumerate}\item A bunch of flowers consists of a mixture of roses, tulips and daffodils. Tom orders a bunch of 7 flowers from a shop to give to a friend. There must be at least 2 of each type of flower. The shop has 6 roses, 5 tulips and 4 daffodils, all different from each other. Find the number of different bunches of flowers that are possible.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2016 Q6 [11]}}