| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Arranging identical items in a line |
| Difficulty | Standard +0.3 This is a multi-part combinatorics question covering standard techniques: arrangements with identical objects (part a), permutations with restrictions (part b), and distribution problems (part c). While it requires careful application of formulas and some logical thinking about constraints, all techniques are routine for S1 level with no novel insights required. Slightly above average due to the variety of methods needed across multiple parts. |
| Spec | 5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{10!}{5! \cdot 4!} = 1260\) | M1 | \(10!\) or \(_{10}P_{10}\) seen in num or alone, or dividing by \(5! \cdot 4!\) only |
| A1 [2] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(_8P_4\) or \(_8C_4 \times 4!\) | M1 | \(_8P_4\) or \(_8C_4\) seen; allow extra multiplication |
| \(= 1680\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6C2 \times 4!\) | M1 | 6C2 or 6P2 seen multiplied; mult by 4! |
| \(= 360\) | M1 | Correct answer |
| OR \(6P4\) or \(4 \times 3 \times 6 \times 5 = 360\) | A1 [3] | Award full marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| A=1, B=1, C=7: \(9C1 \times 8C1 \times 7C7 \times 3C_1 = 216\) | M1 | Summing at least two options of 1,1,7 or 1,3,5 or 3,3,3 |
| A=1, B=3, C=5: \(9C1 \times 8C3 \times 5C5 \times 3! = 3024\) | M1 | Mult an option by 3C1 or 3! or 3C3 |
| A=3, B=3, C=3: \(9C3 \times 6C3 \times 3C3 = 1680\) | M1 | Any one of the \(2^{nd}\) term being \(xCy\) seen; mult fitting with first (\(x\) could be 2,4,5,6 or 8) and corresponding \(y\) |
| A1 | Any of unsimplified 72, 504 or 1680 seen | |
| Total \(= 4920\) ways | A1 [5] | Correct answer |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{10!}{5! \cdot 4!} = 1260$ | M1 | $10!$ or $_{10}P_{10}$ seen in num or alone, or dividing by $5! \cdot 4!$ only |
| | A1 [2] | Correct final answer |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $_8P_4$ or $_8C_4 \times 4!$ | M1 | $_8P_4$ or $_8C_4$ seen; allow extra multiplication |
| $= 1680$ | A1 [2] | Correct answer |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6C2 \times 4!$ | M1 | 6C2 or 6P2 seen multiplied; mult by 4! |
| $= 360$ | M1 | Correct answer |
| OR $6P4$ or $4 \times 3 \times 6 \times 5 = 360$ | A1 [3] | Award full marks |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| A=1, B=1, C=7: $9C1 \times 8C1 \times 7C7 \times 3C_1 = 216$ | M1 | Summing at least two options of 1,1,7 or 1,3,5 or 3,3,3 |
| A=1, B=3, C=5: $9C1 \times 8C3 \times 5C5 \times 3! = 3024$ | M1 | Mult an option by 3C1 or 3! or 3C3 |
| A=3, B=3, C=3: $9C3 \times 6C3 \times 3C3 = 1680$ | M1 | Any one of the $2^{nd}$ term being $xCy$ seen; mult fitting with first ($x$ could be 2,4,5,6 or 8) and corresponding $y$ |
| | A1 | Any of unsimplified 72, 504 or 1680 seen |
| Total $= 4920$ ways | A1 [5] | Correct answer |7
\begin{enumerate}[label=(\alph*)]
\item In a sweet shop 5 identical packets of toffees, 4 identical packets of fruit gums and 9 identical packets of chocolates are arranged in a line on a shelf. Find the number of different arrangements of the packets that are possible if the packets of chocolates are kept together.
\item Jessica buys 8 different packets of biscuits. She then chooses 4 of these packets.
\begin{enumerate}[label=(\roman*)]
\item How many different choices are possible if the order in which Jessica chooses the 4 packets is taken into account?
The 8 packets include 1 packet of chocolate biscuits and 1 packet of custard creams.
\item How many different choices are possible if the order in which Jessica chooses the 4 packets is taken into account and the packet of chocolate biscuits and the packet of custard creams are both chosen?
\end{enumerate}\item 9 different fruit pies are to be divided between 3 people so that each person gets an odd number of pies. Find the number of ways this can be done.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2012 Q7 [12]}}