| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Items NOT together (general separation) |
| Difficulty | Standard +0.3 This is a standard permutations question with repeated letters requiring complementary counting (part a), conditional arrangements (part b), and combinations with constraints (part c). All parts use routine techniques taught in S1 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 | Mark | Guidance |
| \(\frac{10!}{2!4!} - \frac{9!}{4!}\ [75600 - 15120]\) | M1 | \(\frac{10!}{a!b!} - c\), \(a \neq b\), \(a=1,2\), \(b=1,4\), with \(c\) a positive integer |
| M1 | \(d - \frac{e!}{4!}\), \(e = 8,9,10\), with \(d\) a positive integer | |
| \(= 60480\) | A1 | Exact value only. SC B1 for final answer 60480 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{8!}{4!} \times \frac{9 \times 8}{2}\) | M1 | \(\frac{8!}{4!} \times f\), with \(f\) a positive integer |
| M1 | \(g \times \frac{9\times8}{h}\), with \(g\) a positive integer, \(h=1,2\). \(g \times {}^9C_2\) and \(g \times {}^9P_2\) acceptable | |
| \(= 60480\) | A1 | Exact value only. SC B1 for final answer 60480 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AC{\wedge\wedge\wedge}C{\wedge\wedge\wedge}A\): \(\frac{6!}{2!} \times 4\) | M1 | \(\frac{6!}{2!} \times s\), with \(s\) a positive integer |
| M1 | \(\frac{t!}{r!} \times 4\), \(r=1,2,3\) and \(t=8,7,6\) | |
| \(1440\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{4 \times {}^6P_3 \times 3!}{2!}\) | M1 | \(\frac{{}^6P_3}{2!} \times k\), with \(k\) a positive integer |
| M1 | \(4 \times 3! \times \frac{{}^6P_m}{n!}\), \(m=2,3\) and \(n=1,2,3\) | |
| \(1440\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Scenarios: \(AA\_\_\_\): \({}^5C_3 = 10\); \(AAA\_\_\): \({}^5C_2 = 10\); \(AAAA\_\): \({}^5C_1 = 5\) | B1 | Correct number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www |
| M1 | Add 3 values for 2, 3 and 4 As, no additional, incorrect or repeated scenarios. Accept unsimplified | |
| \(25\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Scenarios: \(AAC\_\_\): \({}^4C_2=6\); \(AA\_\_\_\): \({}^4C_3=4\); \(AAAC\_\): \({}^4C_1=4\); \(AAA\_\_\): \({}^4C_2=6\); \(AAAAC\): \(1\); \(AAAA\_\): \(4\) | B1 | Correct total number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www |
| M1 | Add 6 values of appropriate scenarios only, no additional, incorrect or repeated scenarios. Accept unsimplified | |
| \(25\) | A1 |
## Question 7(a):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{10!}{2!4!} - \frac{9!}{4!}\ [75600 - 15120]$ | M1 | $\frac{10!}{a!b!} - c$, $a \neq b$, $a=1,2$, $b=1,4$, with $c$ a positive integer |
| | M1 | $d - \frac{e!}{4!}$, $e = 8,9,10$, with $d$ a positive integer |
| $= 60480$ | A1 | Exact value only. **SC B1** for final answer 60480 www |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{8!}{4!} \times \frac{9 \times 8}{2}$ | M1 | $\frac{8!}{4!} \times f$, with $f$ a positive integer |
| | M1 | $g \times \frac{9\times8}{h}$, with $g$ a positive integer, $h=1,2$. $g \times {}^9C_2$ and $g \times {}^9P_2$ acceptable |
| $= 60480$ | A1 | Exact value only. **SC B1** for final answer 60480 www |
## Question 7(b):
**Main Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $AC{\wedge\wedge\wedge}C{\wedge\wedge\wedge}A$: $\frac{6!}{2!} \times 4$ | M1 | $\frac{6!}{2!} \times s$, with $s$ a positive integer |
| | M1 | $\frac{t!}{r!} \times 4$, $r=1,2,3$ and $t=8,7,6$ |
| $1440$ | A1 | |
**Alternative Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{4 \times {}^6P_3 \times 3!}{2!}$ | M1 | $\frac{{}^6P_3}{2!} \times k$, with $k$ a positive integer |
| | M1 | $4 \times 3! \times \frac{{}^6P_m}{n!}$, $m=2,3$ and $n=1,2,3$ |
| $1440$ | A1 | |
## Question 7(c):
**Main Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Scenarios: $AA\_\_\_$: ${}^5C_3 = 10$; $AAA\_\_$: ${}^5C_2 = 10$; $AAAA\_$: ${}^5C_1 = 5$ | B1 | Correct number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www |
| | M1 | Add 3 values for 2, 3 and 4 As, no additional, incorrect or repeated scenarios. Accept unsimplified |
| $25$ | A1 | |
**Alternative Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Scenarios: $AAC\_\_$: ${}^4C_2=6$; $AA\_\_\_$: ${}^4C_3=4$; $AAAC\_$: ${}^4C_1=4$; $AAA\_\_$: ${}^4C_2=6$; $AAAAC$: $1$; $AAAA\_$: $4$ | B1 | Correct total number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www |
| | M1 | Add 6 values of appropriate scenarios only, no additional, incorrect or repeated scenarios. Accept unsimplified |
| $25$ | A1 | |7
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements of the 10 letters in the word CASABLANCA in which the two Cs are not together.
\item Find the number of different arrangements of the 10 letters in the word CASABLANCA which have an A at the beginning, an A at the end and exactly 3 letters between the 2 Cs .\\
Five letters are selected from the 10 letters in the word CASABLANCA.
\item Find the number of different selections in which the five letters include at least two As and at most one C.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q7 [9]}}