| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Moderate -0.8 This is a straightforward permutations and combinations question testing standard techniques. Part (a) requires basic counting with simple constraints (first digit must be 3 or 4), and part (b) applies routine combination formulas. All parts are direct applications of textbook methods with no novel problem-solving required, making it easier than average for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| First digit in 2 ways. \(2 \times 4 \times 3 \times 2\) or \(2 \times {}^4P3\) | M1 | 1, 2 or \(3 \times {}^4P3\) OE as final answer |
| Total \(= 48\) ways | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2 \times 5 \times 5 \times 3\) | M1 | Seeing \(5^2\) mult; this mark is for correctly considering the middle two digits with replacement |
| M1 | Mult by 6; this mark is for correctly considering the first and last digits | |
| \(= 150\) ways | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(OO\) in \(^{18}C_4\) ways | M1 | \(^{18}C_x\) or the sum of five 2-factor products with \(n=14\) and 4, may be \(\times\) by \(2C2\): \(4C0 \times 14C4 + 4C1 \times 14C3 + 4C2 \times 14C2 + 4C3 \times 14C1 + 4C4 \times 14C0\) |
| \(= 3060\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Choc \(0,1,2\); Not Choc: \(1\times^{16}C_6=8008\), \(^4C_1\times^{16}C_5=17472\), \(^4C_2\times^{16}C_4=10920\) OR table of Choc/Oats/Ginger combinations listed | B1 | The correct number of ways with one of 0, 1 or 2 chocs, unsimplified or any three correct number of ways of combining choc/oat/ginger, unsimplified |
| Total \(= 36400\) ways | M1 | Sum the number of ways with 0, 1 and 2 chocs and two must be totally correct, unsimplified OR sum the nine combinations of choc, ginger, oats, six must be totally correct, unsimplified |
| Probability \(= 36400 / ^{20}C_6\) | M1 | Dividing by \(^{20}C_6\) (38760) |
| \(= 0.939\ (910/969)\) | A1 |
# Question 6(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| First digit in 2 ways. $2 \times 4 \times 3 \times 2$ or $2 \times {}^4P3$ | M1 | 1, 2 or $3 \times {}^4P3$ OE as final answer |
| Total $= 48$ ways | A1 | |
| **Total: 2** | | |
# Question 6(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2 \times 5 \times 5 \times 3$ | M1 | Seeing $5^2$ mult; this mark is for correctly considering the middle two digits with replacement |
| | M1 | Mult by 6; this mark is for correctly considering the first and last digits |
| $= 150$ ways | A1 | |
| **Total: 3** | | |
## Question 6(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $OO****$ in $^{18}C_4$ ways | M1 | $^{18}C_x$ or the sum of five 2-factor products with $n=14$ and 4, may be $\times$ by $2C2$: $4C0 \times 14C4 + 4C1 \times 14C3 + 4C2 \times 14C2 + 4C3 \times 14C1 + 4C4 \times 14C0$ |
| $= 3060$ | A1 | |
---
## Question 6(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Choc $0,1,2$; Not Choc: $1\times^{16}C_6=8008$, $^4C_1\times^{16}C_5=17472$, $^4C_2\times^{16}C_4=10920$ **OR** table of Choc/Oats/Ginger combinations listed | B1 | The correct number of ways with one of 0, 1 or 2 chocs, unsimplified **or** any three correct number of ways of combining choc/oat/ginger, unsimplified |
| Total $= 36400$ ways | M1 | Sum the number of ways with 0, 1 and 2 chocs and two must be totally correct, unsimplified **OR** sum the nine combinations of choc, ginger, oats, six must be totally correct, unsimplified |
| Probability $= 36400 / ^{20}C_6$ | M1 | Dividing by $^{20}C_6$ (38760) |
| $= 0.939\ (910/969)$ | A1 | |
---6
\begin{enumerate}[label=(\alph*)]
\item Find how many numbers between 3000 and 5000 can be formed from the digits $1,2,3,4$ and 5,
\begin{enumerate}[label=(\roman*)]
\item if digits are not repeated,
\item if digits can be repeated and the number formed is odd.
\end{enumerate}\item A box of 20 biscuits contains 4 different chocolate biscuits, 2 different oatmeal biscuits and 14 different ginger biscuits. 6 biscuits are selected from the box at random.
\begin{enumerate}[label=(\roman*)]
\item Find the number of different selections that include the 2 oatmeal biscuits.
\item Find the probability that fewer than 3 chocolate biscuits are selected.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2017 Q6 [11]}}