| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Exactly N letters between items |
| Difficulty | Standard +0.8 This is a multi-part permutations question requiring careful handling of repeated letters (R×2, E×3) and constraints. Parts (a-b) are standard, but part (c) requires systematic counting of positions with exactly 3 letters between Rs, and part (d) involves casework with combinations and repeated letters. The combination of techniques and potential for errors elevates this above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{11!}{2!3!}\) | M1 | 11! alone on numerator – must be a fraction. \(k! \times m!\) on denominator, \(k=1,2\), \(m=1,3,1\) can be implied but cannot both \(= 1\). No additional terms |
| 3326400 | A1 | Exact value only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8! = 40320\) | B1 | Evaluate, exact value only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{9!}{3!} \times 7\) | M1 | \(\frac{9!}{3!} \times k\) seen, \(k\) an integer \(> 0\), no \(+\), \(-\) or \(\div\) |
| M1 | \(7 \times\) an integer seen in final answer, no \(+\), \(-\) or \(\div\) | |
| 423360 | A1 | Exact value only |
| Alternative: \({}^9C_3 \times 7! \times \left(\times\frac{3!}{3!}\right)\) | M1 | \(9C3 \times k\) seen, \(k\) integer \(> 0\) |
| M1 | \(7! \times k\) seen, \(k\) integer \(> 0\) | |
| 423360 | A1 | Exact value, evidence of \(\times\frac{3!}{3!}\) required |
| Alternative: \(3 \times 7 \times \frac{8!}{2!}\) | M1 | \(3 \times \frac{8!}{2!} \times k\) seen, \(k\) integer \(> 0\) |
| M1 | \(7 \times\) an integer in final answer | |
| 423360 | A1 | Exact value only |
| Alternative: \(7 \times \frac{2}{11} \times \frac{9}{10} \times \frac{8}{9} \times \frac{7}{8} \times \frac{1}{7} \times \text{total arrangements}\) | M1 | Product of correct five fractions \(\times k\), \(k\) integer \(> 0\) |
| M1 | \(7 \times \text{total arrangements} \times k\), \(k\) integer \(> 0\) | |
| 423360 | A1 | Exact value only |
| Alternative (complementary): No E between Rs: \(\frac{{}^6C_3 \times 3 \times 7!}{3!} = 100800\); 1E between Rs: \(\frac{{}^6C_2 \times 3 \times 7!}{2!} = 226800\); 2Es between Rs: \({}^6C_1 \times 3 \times 7! = 90720\); 3Es between Rs: \(-7! = 5040\); Total \(= 7\times(20+45+18+1) = 7\times84 = 423360\) | M1 | Finding correct number of ways for 0, 1 or 2 Es between Rs |
| M1 | Adding number of ways for 3 or 4 correct scenarios | |
| 423360 | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Identifying four scenarios: \(EER\_\_\), \(EERR\_\), \(EEER\_\), \(EEERR\) | M1 | Identifying four correct scenarios only |
| \(^6C_2 = 15\), \(^6C_1 = 6\), \(^6C_1 = 6\), \(^6C_0 = 1\) | B1 | Correct number of selections unsimplified for 2 or more scenarios |
| Adding the number of selections for 3 or 4 identified correct scenarios | M1 | \(^3C_x \times ^2C_y \times ^6C_z\), \(x+y+z=5\) correctly identifies \(x\) Es and \(y\) Rs |
| \([\text{Total} =]\ 28\) | A1 | WWW, only dependent upon 2nd M mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(EER\wedge\wedge = {^8C_2}\) | M1 | \(^8C_x\) seen alone or \(^8C_x \times k\), \(k=1\) or \(2\), \(0 < x < 8\); condone \(^8P_x\) or \(^8P_x \times k\) |
| \(^8C_2 \times k\), \(k = 1\) or \(2\) | B1 | |
| \(^8C_2 \times k\), \(k = 1\) OE and no other terms | M1 | |
| \([\text{Total} =]\ 28\) | A1 | Value stated |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{11!}{2!3!}$ | M1 | 11! alone on numerator – must be a fraction. $k! \times m!$ on denominator, $k=1,2$, $m=1,3,1$ can be implied but cannot both $= 1$. No additional terms |
| 3326400 | A1 | Exact value only |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $8! = 40320$ | B1 | Evaluate, exact value only |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{9!}{3!} \times 7$ | M1 | $\frac{9!}{3!} \times k$ seen, $k$ an integer $> 0$, no $+$, $-$ or $\div$ |
| | M1 | $7 \times$ an integer seen in final answer, no $+$, $-$ or $\div$ |
| 423360 | A1 | Exact value only |
| **Alternative:** ${}^9C_3 \times 7! \times \left(\times\frac{3!}{3!}\right)$ | M1 | $9C3 \times k$ seen, $k$ integer $> 0$ |
| | M1 | $7! \times k$ seen, $k$ integer $> 0$ |
| 423360 | A1 | Exact value, evidence of $\times\frac{3!}{3!}$ required |
| **Alternative:** $3 \times 7 \times \frac{8!}{2!}$ | M1 | $3 \times \frac{8!}{2!} \times k$ seen, $k$ integer $> 0$ |
| | M1 | $7 \times$ an integer in final answer |
| 423360 | A1 | Exact value only |
| **Alternative:** $7 \times \frac{2}{11} \times \frac{9}{10} \times \frac{8}{9} \times \frac{7}{8} \times \frac{1}{7} \times \text{total arrangements}$ | M1 | Product of correct five fractions $\times k$, $k$ integer $> 0$ |
| | M1 | $7 \times \text{total arrangements} \times k$, $k$ integer $> 0$ |
| 423360 | A1 | Exact value only |
| **Alternative (complementary):** No E between Rs: $\frac{{}^6C_3 \times 3 \times 7!}{3!} = 100800$; 1E between Rs: $\frac{{}^6C_2 \times 3 \times 7!}{2!} = 226800$; 2Es between Rs: ${}^6C_1 \times 3 \times 7! = 90720$; 3Es between Rs: $-7! = 5040$; Total $= 7\times(20+45+18+1) = 7\times84 = 423360$ | M1 | Finding correct number of ways for 0, 1 or 2 Es between Rs |
| | M1 | Adding number of ways for 3 or 4 correct scenarios |
| 423360 | A1 | CAO |
## Question 6(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Identifying four scenarios: $EER\_\_$, $EERR\_$, $EEER\_$, $EEERR$ | M1 | Identifying four correct scenarios only |
| $^6C_2 = 15$, $^6C_1 = 6$, $^6C_1 = 6$, $^6C_0 = 1$ | B1 | Correct number of selections unsimplified for 2 or more scenarios |
| Adding the number of selections for 3 or 4 identified correct scenarios | M1 | $^3C_x \times ^2C_y \times ^6C_z$, $x+y+z=5$ correctly identifies $x$ Es and $y$ Rs |
| $[\text{Total} =]\ 28$ | A1 | WWW, only dependent upon 2nd M mark |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $EER\wedge\wedge = {^8C_2}$ | M1 | $^8C_x$ seen alone or $^8C_x \times k$, $k=1$ or $2$, $0 < x < 8$; condone $^8P_x$ or $^8P_x \times k$ |
| $^8C_2 \times k$, $k = 1$ or $2$ | B1 | |
| $^8C_2 \times k$, $k = 1$ OE and no other terms | M1 | |
| $[\text{Total} =]\ 28$ | A1 | Value stated |
---6
\begin{enumerate}[label=(\alph*)]
\item How many different arrangements are there of the 11 letters in the word REQUIREMENT? [2]
\item How many different arrangements are there of the 11 letters in the word REQUIREMENT in which the two Rs are together and the three Es are together?
\item How many different arrangements are there of the 11 letters in the word REQUIREMENT in which there are exactly three letters between the two Rs?\\
Five of the 11 letters in the word REQUIREMENT are selected.
\item How many possible selections contain at least two Es and at least one R?
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q6 [10]}}