| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Arrangements with positional constraints |
| Difficulty | Standard +0.3 This is a standard permutations and combinations question with repeated letters (3 As, 2 Ms). Parts (i)-(iii) use the formula n!/r!s! for arrangements with restrictions, while (iv)-(v) are straightforward selection problems. All parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{9!}{2!3!}\) | M1 | 9! alone on numerator, 2! and/or 3! on denominator |
| \(= 30240\) | A1 | Exact value, final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A \wedge\wedge\wedge A \wedge\wedge\wedge A\), Arrangements \(= \frac{6!}{2!} = 360\) | B1 | Final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M \wedge M \wedge\wedge\wedge\wedge\wedge\wedge = \frac{7!}{3!} \times 7\) | M1 | 7! in numerator (considering letters not M) |
| M1 | Division by 3! only (removing repeated As) | |
| M1 | Multiply by 7 (positions of M-M) | |
| \(= 5880\) | A1 | Exact value, final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1\times\frac{6!}{2!}\times7 + 4\times\frac{6!}{3!}\times7\) | M1 | 6! in sum of 2 expressions \(a6! + b6!\) |
| M1 | Multiply by 7 in both expressions (positions of M-M) | |
| \(= 2520 + 3360\) | M1 | \(\frac{c}{2!}+\frac{d}{3!}\) seen (removing repeated As) |
| \(= 5880\) | A1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((MAM)\wedge\wedge\wedge\wedge\wedge\wedge = 7!/2! = 2520\) | M1 | 7! in numerator (considering 6 letters + block) |
| \((MA'M)\wedge\wedge\wedge\wedge\wedge\wedge = 7!/3! \times 4 = 840 \times 4 = 3360\) | M1 | Division by 2! and 3! seen in different terms |
| \(\text{Total} = 2520 + 3360\) | M1 | Summing 5 correct scenarios only |
| \(= 5880\) | A1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(MA\wedge = {}^4C_1 = 4\) | B1 | Final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M\wedge\wedge: {}^4C_2 = 6\), \(MM\wedge: {}^4C_1 = 4\) | M1 | Either option \(MM\wedge\) or \(M\wedge\wedge\) correct, accept unsimplified |
| \(MMA = 1\), \(MAA = 1\), \((MA\_: {}^4C_1 = 4)\) | M1 | Add 4 or 5 correct scenarios only |
| \(\text{Total} = 16\) | A1 | Value must be clearly stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(MM\wedge = {}^5C_1 = 5\) | M1 | Either option \(MM\wedge\) or \(M\wedge\wedge\) correct, accept unsimplified |
| \(M\wedge\wedge = {}^5C_2 = 10\) | M1 | Adding 2 or 3 correct scenarios only |
| \(MAA = 1\), \(\text{Total} = 16\) | A1 | Value must be clearly stated |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{9!}{2!3!}$ | M1 | 9! alone on numerator, 2! and/or 3! on denominator |
| $= 30240$ | A1 | Exact value, final answer |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A \wedge\wedge\wedge A \wedge\wedge\wedge A$, Arrangements $= \frac{6!}{2!} = 360$ | B1 | Final answer |
---
## Question 7(iii):
**Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M \wedge M \wedge\wedge\wedge\wedge\wedge\wedge = \frac{7!}{3!} \times 7$ | M1 | 7! in numerator (considering letters not M) |
| | M1 | Division by 3! only (removing repeated As) |
| | M1 | Multiply by 7 (positions of M-M) |
| $= 5880$ | A1 | Exact value, final answer |
**Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1\times\frac{6!}{2!}\times7 + 4\times\frac{6!}{3!}\times7$ | M1 | 6! in sum of 2 expressions $a6! + b6!$ |
| | M1 | Multiply by 7 in both expressions (positions of M-M) |
| $= 2520 + 3360$ | M1 | $\frac{c}{2!}+\frac{d}{3!}$ seen (removing repeated As) |
| $= 5880$ | A1 | Exact value |
**Method 3:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(MAM)\wedge\wedge\wedge\wedge\wedge\wedge = 7!/2! = 2520$ | M1 | 7! in numerator (considering 6 letters + block) |
| $(MA'M)\wedge\wedge\wedge\wedge\wedge\wedge = 7!/3! \times 4 = 840 \times 4 = 3360$ | M1 | Division by 2! and 3! seen in different terms |
| $\text{Total} = 2520 + 3360$ | M1 | Summing 5 correct scenarios only |
| $= 5880$ | A1 | Exact value |
---
## Question 7(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $MA\wedge = {}^4C_1 = 4$ | B1 | Final answer |
---
## Question 7(v):
**Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M\wedge\wedge: {}^4C_2 = 6$, $MM\wedge: {}^4C_1 = 4$ | M1 | Either option $MM\wedge$ or $M\wedge\wedge$ correct, accept unsimplified |
| $MMA = 1$, $MAA = 1$, $(MA\_: {}^4C_1 = 4)$ | M1 | Add 4 or 5 correct scenarios only |
| $\text{Total} = 16$ | A1 | Value must be clearly stated |
**Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $MM\wedge = {}^5C_1 = 5$ | M1 | Either option $MM\wedge$ or $M\wedge\wedge$ correct, accept unsimplified |
| $M\wedge\wedge = {}^5C_2 = 10$ | M1 | Adding 2 or 3 correct scenarios only |
| $MAA = 1$, $\text{Total} = 16$ | A1 | Value must be clearly stated |7 Find the number of different arrangements that can be made of all 9 letters in the word CAMERAMAN in each of the following cases.\\
(i) There are no restrictions.\\
(ii) The As occupy the 1st, 5th and 9th positions.\\
(iii) There is exactly one letter between the Ms.\\
Three letters are selected from the 9 letters of the word CAMERAMAN.\\
(iv) Find the number of different selections if the three letters include exactly one M and exactly one A.\\
(v) Find the number of different selections if the three letters include at least one M.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S1 2019 Q7 [11]}}