| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.3 This is a multi-part combinatorics problem requiring systematic counting across three scenarios. Part (a) involves conditional selection with constraints, part (b) adds seating arrangement considerations, and part (c) is a standard permutation with adjacency constraints. While it requires careful case analysis and multiple techniques (combinations, permutations, restrictions), each individual step uses routine A-level methods without requiring novel insight. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(^5C_2 \times 2\) | M1 | \(^5C_2 \times r\), \(r\) = positive integer, 1 implied, no addition |
| M1 | \(s \times 2\), \(s = {^5C_2}\) or \(^5P_2\); or if \(^5C_2\) or \(^5P_2\) not present, \(s\) = single integer \(> 1\); or \(t! \times 2\), \(2 \leqslant t \leqslant 8\), no other terms | |
| \(20\) | A1 | |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \(^6C_2 \times 2 \times 2\times 2 \times 4!\) | M1 | \(^6C_2 \times 2 \times 2\times 2 \times t\), \(t\) = positive integer \(\geqslant 1\); or \(^6P_2 \times 2 \times 2 \times t\) |
| M1 | \(u \times 4!\), \(u\) = positive integer \(> 1\) | |
| \(2880\) | A1 | If A0 scored, SC B1 for \(2880\) nfww |
| Method 2: \(6! \times 2 \times 2\) | M1 | \(6! \times v\), \(v\) = positive integer \(\geqslant 1\) |
| M1 | \(w \times 2 \times 2\), \(w\) = positive integer \(> 1\); condone \(w \times 4\) | |
| \(2880\) | A1 | If A0 scored, SC B1 for \(2880\) nfww |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(7! \times 2 - 6! \times 4\) | M1 | \(7! \times 2 - a\), \(a\) = positive integer \(> 1\) |
| M1 | \(b - 6! \times 4\), \(b\) = positive integer \(> 2880\) | |
| M1 | \(7! \times c - 6! \times d\), \(c = 1, 2\) and \(d = 1, 4\) | |
| \(= 7200\) | A1 | If A0 scored, SC B1 for 7200 nfww |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6! \times 2 \times 5\) | M1 | \(6! \times e \times f\), \(e, f\) = positive integers \(\geqslant 1\) |
| M1 | \(6! \times 2 \times f\), \(f\) = positive integer \(\geqslant 1\). If 5! used, SC B1 \(5! \times 2 \times f\), \(f\) = positive integer \(> 1\) | |
| M1 | \(6! \times e \times 5\), \(e\) = positive integer \(\geqslant 1\) | |
| \(7200\) | A1 | If A0, scored SC B1 for 7200 nfww |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(^6P_2 \times 5! \times 2\) | M1 | \(^6P_2 \times e \times f\), \(e, f\) = positive integers \(\geqslant 1\) |
| M1 | \(^6P_2 \times 5! \times f\), \(f\) = positive integer \(\geqslant 1\). Condone \(^6C_2 \times 5! \times f\), \(f\) = positive integer \(\geqslant 1\) | |
| M1 | \(^6P_2 \times e \times 2\), \(e\) = positive integer \(\geqslant 1\). Condone \(^6C_2 \times e \times 2\), \(e\) = positive integer \(\geqslant 1\) | |
| \(7200\) | A1 | If A0 scored, SC B1 for 7200 nfww |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\wedge RS\wedge\wedge\wedge\wedge\wedge\): \(^6P_1 \times 2 \times 5! = 1440\) | M1 | \(^6P_n \times a \times (6-n)!\), \(a\) = positive integer, \(1 \leqslant n \leqslant 5\) seen once |
| \(\wedge\wedge RS\wedge\wedge\wedge\wedge\): \(^6P_2 \times 2 \times 4! = 1440\) | ||
| \(\wedge\wedge\wedge RS\wedge\wedge\wedge\): \(^6P_3 \times 2 \times 3! = 1440\) | M1 | \(^6P_n \times 2 \times (6-n)!\), \(a\) = positive integer, \(1 \leqslant n \leqslant 5\) seen at least 3 times in identified scenarios |
| \(\wedge\wedge\wedge\wedge RS\wedge\wedge\): \(^6P_4 \times 2 \times 2! = 1440\) | ||
| \(\wedge\wedge\wedge\wedge\wedge RS\wedge\): \(^6P_5 \times 2 \times 1! = 1440\) | M1 | Add 5 values of appropriate scenarios only. No additional, incorrect or repeated scenarios. Accept unsimplified |
| \(7200\) | A1 | If A0 scored, SC B1 for 7200 nfww |
| [4] |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $^5C_2 \times 2$ | **M1** | $^5C_2 \times r$, $r$ = positive integer, 1 implied, no addition |
| | **M1** | $s \times 2$, $s = {^5C_2}$ or $^5P_2$; or if $^5C_2$ or $^5P_2$ not present, $s$ = single integer $> 1$; or $t! \times 2$, $2 \leqslant t \leqslant 8$, no other terms |
| $20$ | **A1** | |
| **Total: 3 marks** | | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $^6C_2 \times 2 \times 2\times 2 \times 4!$ | **M1** | $^6C_2 \times 2 \times 2\times 2 \times t$, $t$ = positive integer $\geqslant 1$; or $^6P_2 \times 2 \times 2 \times t$ |
| | **M1** | $u \times 4!$, $u$ = positive integer $> 1$ |
| $2880$ | **A1** | If A0 scored, **SC B1** for $2880$ nfww |
| **Method 2:** $6! \times 2 \times 2$ | **M1** | $6! \times v$, $v$ = positive integer $\geqslant 1$ |
| | **M1** | $w \times 2 \times 2$, $w$ = positive integer $> 1$; condone $w \times 4$ |
| $2880$ | **A1** | If A0 scored, **SC B1** for $2880$ nfww |
| **Total: 3 marks** | | |
## Question 6(c):
**Method 1:** Number of arrangements with Rajid and Sue together – Number of arrangements with Rajid and Sue together and at end of line
| Answer/Working | Mark | Guidance |
|---|---|---|
| $7! \times 2 - 6! \times 4$ | **M1** | $7! \times 2 - a$, $a$ = positive integer $> 1$ |
| | **M1** | $b - 6! \times 4$, $b$ = positive integer $> 2880$ |
| | **M1** | $7! \times c - 6! \times d$, $c = 1, 2$ and $d = 1, 4$ |
| $= 7200$ | **A1** | If A0 scored, **SC B1** for 7200 nfww |
**Method 2:** Arrangements of 6 people and then place Rajid and Sue
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6! \times 2 \times 5$ | **M1** | $6! \times e \times f$, $e, f$ = positive integers $\geqslant 1$ |
| | **M1** | $6! \times 2 \times f$, $f$ = positive integer $\geqslant 1$. If 5! used, **SC B1** $5! \times 2 \times f$, $f$ = positive integer $> 1$ |
| | **M1** | $6! \times e \times 5$, $e$ = positive integer $\geqslant 1$ |
| $7200$ | **A1** | If A0, scored **SC B1** for 7200 nfww |
**Method 3:** Friends at ends picked first $F \wedge RS \wedge\wedge\wedge F$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $^6P_2 \times 5! \times 2$ | **M1** | $^6P_2 \times e \times f$, $e, f$ = positive integers $\geqslant 1$ |
| | **M1** | $^6P_2 \times 5! \times f$, $f$ = positive integer $\geqslant 1$. Condone $^6C_2 \times 5! \times f$, $f$ = positive integer $\geqslant 1$ |
| | **M1** | $^6P_2 \times e \times 2$, $e$ = positive integer $\geqslant 1$. Condone $^6C_2 \times e \times 2$, $e$ = positive integer $\geqslant 1$ |
| $7200$ | **A1** | If A0 scored, **SC B1** for 7200 nfww |
**Method 4:** RS placed in different possible positions
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\wedge RS\wedge\wedge\wedge\wedge\wedge$: $^6P_1 \times 2 \times 5! = 1440$ | **M1** | $^6P_n \times a \times (6-n)!$, $a$ = positive integer, $1 \leqslant n \leqslant 5$ seen once |
| $\wedge\wedge RS\wedge\wedge\wedge\wedge$: $^6P_2 \times 2 \times 4! = 1440$ | | |
| $\wedge\wedge\wedge RS\wedge\wedge\wedge$: $^6P_3 \times 2 \times 3! = 1440$ | **M1** | $^6P_n \times 2 \times (6-n)!$, $a$ = positive integer, $1 \leqslant n \leqslant 5$ seen at least 3 times in identified scenarios |
| $\wedge\wedge\wedge\wedge RS\wedge\wedge$: $^6P_4 \times 2 \times 2! = 1440$ | | |
| $\wedge\wedge\wedge\wedge\wedge RS\wedge$: $^6P_5 \times 2 \times 1! = 1440$ | **M1** | Add 5 values of appropriate scenarios only. No additional, incorrect or repeated scenarios. Accept unsimplified |
| $7200$ | **A1** | If A0 scored, **SC B1** for 7200 nfww |
| | **[4]** | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{e8c2b51e-d788-4917-829e-1b056a24f520-12_291_809_255_667}
In a restaurant, the tables are rectangular. Each table seats four people: two along each of the longer sides of the table (see diagram). Eight friends have booked two tables, $X$ and $Y$. Rajid, Sue and Tan are three of these friends.
\begin{enumerate}[label=(\alph*)]
\item The eight friends will be divided into two groups of 4, one group for table $X$ and one group for table $Y$.
Find the number of ways in which this can be done if Rajid and Sue must sit at the same table as each other and Tan must sit at the other table.\\
When the friends arrive at the restaurant, Rajid and Sue now decide to sit at table $X$ on the same side as each other. Tan decides that he does not mind at which table he sits.
\item Find the number of different seating arrangements for the 8 friends.\\
As they leave the restaurant, the 8 friends stand in a line for a photograph.
\item Find the number of different arrangements if Rajid and Sue stand next to each other, but neither is at an end of the line.\\
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 Q6 [10]}}