| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.3 This is a standard combinations question with straightforward restrictions. Part (a) requires splitting into two cases (both in/both out) and adding - a routine technique. Part (b) involves permutations with common restrictions (complementary counting for 'not adjacent', treating groups as single units). All methods are textbook exercises requiring no novel insight, though the multi-part nature and careful case-work place it slightly above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Boys in: \({}_{10}C_1 \times {}_{9}C_3 = 840\) ways | M1 | Summing two 2-factor products, C or P |
| Boys out: \({}_{10}C_3 \times {}_{9}C_3 = 10080\) ways | B1 | Any correct option unsimplified |
| Total \(= 10920\) ways \((10900)\) | A1 [3] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \({}_{12}P_8 = 19{,}958{,}400\) | B1 [1] | or 20,000,000 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Together: \({}_{11}P_7 = 1663200 \times 2 = 3326400\) | B1 | \({}_{11}P_7\) seen |
| Not together: \(19958400 - 3326400\) | M1 | \(19958400\) or their (i) \(-\) their together (must be \(> 0\)) |
| \(= 16{,}632{,}000\ (16{,}600{,}000)\) | A1 [3] | Correct final answer |
| OR M at end then not F in \(10 \times {}_{10}P_6 \times 2 = 3024000\) ways | M1 | Summing options for M at end and M not at end |
| Not at end in \(10 \times 9 \times {}_{10}P_6 = 13608000\) ways | B1 | One correct option |
| Total \(= 16{,}632{,}000\) ways | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8! \times 5 = 201600\) ways | B1 | \(8!\) seen multiplied by equivalent of integer \(\geq 1\) |
| M1 | Multiply by 5 | |
| A1 [3] | Correct answer. SR: \(8! \times 5! = 4838400\): B2 |
## Question 5:
**(a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Boys in: ${}_{10}C_1 \times {}_{9}C_3 = 840$ ways | M1 | Summing two 2-factor products, C or P |
| Boys out: ${}_{10}C_3 \times {}_{9}C_3 = 10080$ ways | B1 | Any correct option unsimplified |
| Total $= 10920$ ways $(10900)$ | A1 [3] | Correct final answer |
**(b)(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| ${}_{12}P_8 = 19{,}958{,}400$ | B1 [1] | or 20,000,000 |
**(b)(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Together: ${}_{11}P_7 = 1663200 \times 2 = 3326400$ | B1 | ${}_{11}P_7$ seen |
| Not together: $19958400 - 3326400$ | M1 | $19958400$ or their (i) $-$ their together (must be $> 0$) |
| $= 16{,}632{,}000\ (16{,}600{,}000)$ | A1 [3] | Correct final answer |
| **OR** M at end then not F in $10 \times {}_{10}P_6 \times 2 = 3024000$ ways | M1 | Summing options for M at end and M not at end |
| Not at end in $10 \times 9 \times {}_{10}P_6 = 13608000$ ways | B1 | One correct option |
| Total $= 16{,}632{,}000$ ways | A1 | Correct final answer |
**(b)(iii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $8! \times 5 = 201600$ ways | B1 | $8!$ seen multiplied by equivalent of integer $\geq 1$ |
| | M1 | Multiply by 5 |
| | A1 [3] | Correct answer. SR: $8! \times 5! = 4838400$: B2 |
---5
\begin{enumerate}[label=(\alph*)]
\item A team of 3 boys and 3 girls is to be chosen from a group of 12 boys and 9 girls to enter a competition. Tom and Henry are two of the boys in the group. Find the number of ways in which the team can be chosen if Tom and Henry are either both in the team or both not in the team.
\item The back row of a cinema has 12 seats, all of which are empty. A group of 8 people, including Mary and Frances, sit in this row. Find the number of different ways they can sit in these 12 seats if
\begin{enumerate}[label=(\roman*)]
\item there are no restrictions,
\item Mary and Frances do not sit in seats which are next to each other,
\item all 8 people sit together with no empty seats between them.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2012 Q5 [10]}}