| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | November |
| 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 with standard scenarios. Part (a) involves basic counting principles with simple constraints (first digit fixed as 5), and part (b) uses routine combination formulas with case-by-case enumeration. All techniques are textbook exercises requiring only methodical application of formulas rather than problem-solving insight. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1\times5\times4\times3\) or \({}^5C_3\times3!\) or \({}^5P_3\) | M1 | One of these oe |
| \(= 60\) | A1 [2] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1\times6^3 = 216\) | M1 | Seeing \(6^3\) |
| \(= 216\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5G\ 0B = {}^8C_5 = 56\ (\times{}^6C_0)\) | M1 | \(\Sigma\) 2 or three 2-factor products, C or P |
| \(4G\ 1B = {}^8C_4\times{}^6C_1 = 420\) | B1 | Any correct option unsimplified |
| \(3G\ 2B = {}^8C_3\times{}^6C_2 = 840\) | A1 | A second correct option unsimplified |
| total \(= 1316\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \({}^{11}C_2 + {}^{11}C_5\) | M1 | Adding two single perm or comb options \({}^{11}C_x + {}^{11}C_y\) |
| \(= 55 + 462\) | B1 | One correct unsimplified option |
| \(= 517\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(3B,2G)+P(4B,1G)+P(5B,0G)+\text{cousins out}\ P(3B,2G)\) \(+P(2B,3G)+P(1B,4G)+P(0B,5G)\) \(= 28+24+3+28+168+210+56 = 517\) | M1 | \(\Sigma\) 5 or more 2-factor perm or comb terms |
| B1 | 3 or more correct unsimplified options | |
| \(= 517\) | A1 [3] | Correct answer |
# Question 5:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1\times5\times4\times3$ or ${}^5C_3\times3!$ or ${}^5P_3$ | M1 | One of these oe |
| $= 60$ | A1 **[2]** | Correct final answer |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1\times6^3 = 216$ | M1 | Seeing $6^3$ |
| $= 216$ | A1 **[2]** | Correct answer |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5G\ 0B = {}^8C_5 = 56\ (\times{}^6C_0)$ | M1 | $\Sigma$ 2 or three 2-factor products, C or P |
| $4G\ 1B = {}^8C_4\times{}^6C_1 = 420$ | B1 | Any correct option unsimplified |
| $3G\ 2B = {}^8C_3\times{}^6C_2 = 840$ | A1 | A second correct option unsimplified |
| total $= 1316$ | A1 **[4]** | Correct answer |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| ${}^{11}C_2 + {}^{11}C_5$ | M1 | Adding two single perm or comb options ${}^{11}C_x + {}^{11}C_y$ |
| $= 55 + 462$ | B1 | One correct unsimplified option |
| $= 517$ | A1 **[3]** | Correct answer |
**OR:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(3B,2G)+P(4B,1G)+P(5B,0G)+\text{cousins out}\ P(3B,2G)$ $+P(2B,3G)+P(1B,4G)+P(0B,5G)$ $= 28+24+3+28+168+210+56 = 517$ | M1 | $\Sigma$ 5 or more 2-factor perm or comb terms |
| | B1 | 3 or more correct unsimplified options |
| $= 517$ | A1 **[3]** | Correct answer |
---5
\begin{enumerate}[label=(\alph*)]
\item Find how many numbers between 5000 and 6000 can be formed from the digits 1, 2, 3, 4, 5 and 6
\begin{enumerate}[label=(\roman*)]
\item if no digits are repeated,
\item if repeated digits are allowed.
\end{enumerate}\item Find the number of ways of choosing a school team of 5 pupils from 6 boys and 8 girls
\begin{enumerate}[label=(\roman*)]
\item if there are more girls than boys in the team,
\item if three of the boys are cousins and are either all in the team or all not in the team.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2009 Q5 [11]}}