| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.8 This is a multi-part combinatorics problem requiring careful case analysis in part (a), understanding of multinomial coefficients vs arrangements in part (b), and inclusion-exclusion with constraints in part (c). The progression from selection to arrangement with multiple simultaneous constraints elevates this above standard textbook exercises, though the individual techniques are A-level appropriate. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 5M0W \(^8C_5 \times {}^7C_0 = 56\) | M1 | \(^8C_x \times {}^7C_{5-x}\) for \(x = 1, 2, 3, 4,\) or \(5\) |
| 4M1W \(^8C_4 \times {}^7C_1 = 490\) | B1 | Outcome for 4M1W or 3M2W correct and identified, accept unsimplified |
| 3M2W \(^8C_3 \times {}^7C_2 = 1176\) | M1 | Add 3 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified. Addition may be implied by final answer |
| [Total =] 1722 | A1 | Value stated WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 2M3W \(^8C_2 \times {}^7C_3 = 980\) | M1 | \(^8C_x \times {}^7C_{5-x}\) for \(x = 1, 2, 3, 4,\) or \(5\) |
| 1M4W \(^8C_1 \times {}^7C_4 = 280\) | B1 | Outcome for 2M3W or 1M4W correct and identified, accept unsimplified |
| 0M5W \(^8C_0 \times {}^7C_5 = 21\) | ||
| [Total \(= {}^{15}C_5 - (980 + 280 + 21)\)] \(3003 - (980 + 280 + 21)\) | M1 | Subtract 3 values of appropriate scenarios from *their* identified total or correct, no incorrect scenarios, no repeated scenarios, accept unsimplified |
| [Total =] 1722 | A1 | Value stated WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(^{15}C_3 \times {}^{12}C_5 \times {}^7C_7\ [= 455 \times 792]\) | M1 | \(^{15}C_r \times q\), \(r = 3, 5, 7\); \(q\) a positive integer \(>1\) |
| M1 | \(^{15}C_s \times {}^{15-s}C_t \times {}^{15-s-t}C_u\); \(s = 3,5,7\); \(t = 3,5,7 \neq s\); \(u = 3,5,7 \neq s,t\) | |
| 360360 | A1 | Final answer. If A0 awarded SC B1 for final answer 360360 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6! \times 2 - 5! \times 2 \times 2\ [= 1440 - 480]\) | M1 | \(a! \times 2! \times b\), \(a = 5, 6\); \(b = 1,2\) seen |
| M1 | Either \(6! \times 2 - c\), \(1 < c < 1440\) or \(d - 5! \times 2 \times 2\), \(1440 < d\) | |
| 960 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2 \times 4! \times 5 \times 4\) | M1 | \(2 \times 4! \times e\), \(e\) positive integer \(>1\) |
| M1 | \(f \times 5 \times 4\), \(f\) positive integer \(>1\); condone \(f \times 20\), \(f \times {}^5C_2\), \(f\) positive integer \(>1\) | |
| 960 | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| 5M0W $^8C_5 \times {}^7C_0 = 56$ | M1 | $^8C_x \times {}^7C_{5-x}$ for $x = 1, 2, 3, 4,$ or $5$ |
| 4M1W $^8C_4 \times {}^7C_1 = 490$ | B1 | Outcome for 4M1W **or** 3M2W correct and identified, accept unsimplified |
| 3M2W $^8C_3 \times {}^7C_2 = 1176$ | M1 | Add 3 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified. Addition may be implied by final answer |
| [Total =] 1722 | A1 | Value stated WWW |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| 2M3W $^8C_2 \times {}^7C_3 = 980$ | M1 | $^8C_x \times {}^7C_{5-x}$ for $x = 1, 2, 3, 4,$ or $5$ |
| 1M4W $^8C_1 \times {}^7C_4 = 280$ | B1 | Outcome for 2M3W **or** 1M4W correct and identified, accept unsimplified |
| 0M5W $^8C_0 \times {}^7C_5 = 21$ | | |
| [Total $= {}^{15}C_5 - (980 + 280 + 21)$] $3003 - (980 + 280 + 21)$ | M1 | Subtract 3 values of appropriate scenarios from *their* identified total or correct, no incorrect scenarios, no repeated scenarios, accept unsimplified |
| [Total =] 1722 | A1 | Value stated WWW |
**Total: 4 marks**
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $^{15}C_3 \times {}^{12}C_5 \times {}^7C_7\ [= 455 \times 792]$ | M1 | $^{15}C_r \times q$, $r = 3, 5, 7$; $q$ a positive integer $>1$ |
| | M1 | $^{15}C_s \times {}^{15-s}C_t \times {}^{15-s-t}C_u$; $s = 3,5,7$; $t = 3,5,7 \neq s$; $u = 3,5,7 \neq s,t$ |
| 360360 | A1 | Final answer. If A0 awarded **SC B1** for final answer 360360 |
**Total: 3 marks**
---
## Question 6(c):
**Method 1: Total arrangements with AB together − Arrangements with AB and FG together**
| Answer | Mark | Guidance |
|--------|------|----------|
| $6! \times 2 - 5! \times 2 \times 2\ [= 1440 - 480]$ | M1 | $a! \times 2! \times b$, $a = 5, 6$; $b = 1,2$ seen |
| | M1 | Either $6! \times 2 - c$, $1 < c < 1440$ or $d - 5! \times 2 \times 2$, $1440 < d$ |
| 960 | A1 | |
**Method 2: Arrangements with AB together with F and G not together**
| Answer | Mark | Guidance |
|--------|------|----------|
| $2 \times 4! \times 5 \times 4$ | M1 | $2 \times 4! \times e$, $e$ positive integer $>1$ |
| | M1 | $f \times 5 \times 4$, $f$ positive integer $>1$; condone $f \times 20$, $f \times {}^5C_2$, $f$ positive integer $>1$ |
| 960 | A1 | |
**Total: 3 marks**6 A Social Club has 15 members, of whom 8 are men and 7 are women. The committee of the club consists of 5 of its members.
\begin{enumerate}[label=(\alph*)]
\item Find the number of different ways in which the committee can be formed from the 15 members if it must include more men than women.\\
The 15 members are having their photograph taken. They stand in three rows, with 3 people in the front row, 5 people in the middle row and 7 people in the back row.
\item In how many different ways can the 15 members of the club be divided into a group of 3, a group of 5 and a group of 7 ?\\
In one photograph Abel, Betty, Cally, Doug, Eve, Freya and Gino are the 7 members in the back row.
\item In how many different ways can these 7 members be arranged so that Abel and Betty are next to each other and Freya and Gino are not next to each other?\\
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 2022 Q6 [10]}}