| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with identical objects |
| Difficulty | Standard +0.8 This is a multi-part permutations problem requiring careful case analysis and the constraint that identical objects must be handled correctly. Part (i) needs casework for different end-book scenarios with conditional counting, while part (ii) requires treating grouped objects as units combined with ensuring non-adjacent placement of B books—both requiring systematic problem-solving beyond routine formula application. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| *EITHER* Route 1: \(A\) in \(\frac{9!}{2!2!5!} = 756\) ways | (*M1 | Considering AA and BB options with values |
| \(B\) in \(\frac{9!}{4!5!} = 126\) ways | A1 | Any one option correct |
| *OR* Route 2: \(A\) in \({}^9C_5 \times {}^4C_2 = 756\) ways | (M1 | Considering AA and BB options with values |
| \(B\) in \({}^9C_4 \times {}^5C_5 = 126\) ways | A1 | Any one option correct |
| \(756 + 126\) | DM1 | Summing their AA and BB outcomes only |
| Total \(= 882\) ways | A1 | |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| *EITHER* (subtraction method): As together, no restrictions \(\frac{8!}{2!5!} = 168\) | (*M1 | Considering all As together – 8! seen alone or as numerator; condone \(\times 4!\) for thinking A's not identical |
| As together and Bs together \(\frac{7!}{5!} = 42\) | M1 | Considering all As together and all Bs together – 7! seen alone or numerator |
| M1 | Removing repeated Bs or Cs – dividing by 5! either expression or 2! 1st expression only | |
| Total \(168 - 42\) | DM1 | Subtract their 42 from their 168 (dependent upon first M being awarded) |
| \(= 126\) | A1 | |
| *OR1*: As together, no restrictions \({}^8C_5 \times {}^3C_1 = 168\) | (*M1 | \({}^8C_5\) seen alone or multiplied |
| M1 | \({}^7C_5\) seen alone or multiplied | |
| As together and Bs together \({}^7C_5 \times {}^2C_1 = 42\) | M1 | First expression \(\times {}^3C_1\) or second expression \(\times {}^2C_1\) |
| Total \(168 - 42\) | DM1 | Subtract their 42 from their 168 |
| \(= 126\) | A1 | |
| *OR2* (intersperse method): \((AAAA)CCCCC\) then intersperse \(B\) and another \(B\) | (M1 | Considering all "As together" with Cs – multiply by 6! |
| M1 | Removing repeated Cs – dividing by 5! | |
| *M1 | Considering positions for Bs – multiply by 7P2 | |
| \(\frac{6!}{5!} \times 7 \times 6 \div 2\) | DM1 | Dividing by 2! – removing repeated Bs (dependent upon 3rd M being awarded) |
| \(= 126\) | A1 | |
| Total: | 5 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| *EITHER* Route 1: $A$ in $\frac{9!}{2!2!5!} = 756$ ways | (*M1 | Considering AA and BB options with values |
| $B$ in $\frac{9!}{4!5!} = 126$ ways | A1 | Any one option correct |
| *OR* Route 2: $A$ in ${}^9C_5 \times {}^4C_2 = 756$ ways | (M1 | Considering AA and BB options with values |
| $B$ in ${}^9C_4 \times {}^5C_5 = 126$ ways | A1 | Any one option correct |
| $756 + 126$ | DM1 | Summing their AA and BB outcomes only |
| Total $= 882$ ways | A1 | |
| **Total:** | **4** | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| *EITHER* (subtraction method): As together, no restrictions $\frac{8!}{2!5!} = 168$ | (*M1 | Considering all As together – 8! seen alone or as numerator; condone $\times 4!$ for thinking A's not identical |
| As together and Bs together $\frac{7!}{5!} = 42$ | M1 | Considering all As together and all Bs together – 7! seen alone or numerator |
| | M1 | Removing repeated Bs or Cs – dividing by 5! either expression or 2! 1st expression only |
| Total $168 - 42$ | DM1 | Subtract their 42 from their 168 (dependent upon first M being awarded) |
| $= 126$ | A1 | |
| *OR1*: As together, no restrictions ${}^8C_5 \times {}^3C_1 = 168$ | (*M1 | ${}^8C_5$ seen alone or multiplied |
| | M1 | ${}^7C_5$ seen alone or multiplied |
| As together and Bs together ${}^7C_5 \times {}^2C_1 = 42$ | M1 | First expression $\times {}^3C_1$ or second expression $\times {}^2C_1$ |
| Total $168 - 42$ | DM1 | Subtract their 42 from their 168 |
| $= 126$ | A1 | |
| *OR2* (intersperse method): $(AAAA)CCCCC$ then intersperse $B$ and another $B$ | (M1 | Considering all "As together" with Cs – multiply by 6! |
| | M1 | Removing repeated Cs – dividing by 5! |
| | *M1 | Considering positions for Bs – multiply by 7P2 |
| $\frac{6!}{5!} \times 7 \times 6 \div 2$ | DM1 | Dividing by 2! – removing repeated Bs (dependent upon 3rd M being awarded) |
| $= 126$ | A1 | |
| **Total:** | **5** | |6 A library contains 4 identical copies of book $A , 2$ identical copies of book $B$ and 5 identical copies of book $C$. These 11 books are arranged on a shelf in the library.\\
(i) Calculate the number of different arrangements if the end books are either both book $A$ or both book $B$.\\
(ii) Calculate the number of different arrangements if all the books $A$ are next to each other and none of the books $B$ are next to each other.\\
\hfill \mbox{\textit{CAIE S1 2017 Q6 [9]}}