| Exam Board | OCR |
|---|---|
| Module | Further Discrete (Further Discrete) |
| Year | 2018 |
| Session | March |
| Marks | 8 |
| Topic | Principle of Inclusion/Exclusion |
| Type | Finding Set Cardinalities from Constraints |
| Difficulty | Standard +0.8 This is a multi-part Further Maths discrete question requiring three-set Venn diagram reasoning, pigeonhole principle application, and combinatorial counting with constraints. Part (i) involves careful set theory manipulation across three categories, part (ii) requires understanding pigeonhole principle in a seating context, and part (iii) demands recognizing that each person can appear in at most 3 rounds (since once paired with 2 others, they're exhausted for that set), leading to systematic counting. While the individual concepts are A-level accessible, the combination and the constraint-based reasoning in part (iii) elevates this above typical questions. |
| Spec | 7.01b Set notation: basic language and notation of sets, partitions7.01c Pigeonhole principle7.01j Selections with several constraints |
| Answer | Marks | Guidance |
|---|---|---|
| \(50 - 15 = 21 + 22 + 20 - n(G \cap W) - n(W \cap L) - n(L \cap G) + 2\). \(n(G \cap W) + n(W \cap L) + n(L \cap G) = 63 + 2 - 35 = 30\). \(n(\text{exactly 2 categories}) = n(G \cap W) - 2 + n(W \cap L) - 2 + n(L \cap G) - 2 = 24\) | M1, M1 ft, A1 | Inclusion-exclusion with 3 sets or a 3-set Venn diagram; 63, 2 and 35; 24 as final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Each row = 20 seats = 10 pairs. There must be (at least) 11 of the people in one row of 10 pairs. So at least one pair with both seats used. | M1, A1 | Using pairs of seats; Using pigeonhole principle |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. ABC ADE AFG BDF BEG CDG CEF = 7 rounds. \(7 + 7 + 7 = 21\) | M1, A1, E1 ft | Listing triples; 7 (rounds/triples), cao; \(3 \times\) their 7 |
## (i)
| $50 - 15 = 21 + 22 + 20 - n(G \cap W) - n(W \cap L) - n(L \cap G) + 2$. $n(G \cap W) + n(W \cap L) + n(L \cap G) = 63 + 2 - 35 = 30$. $n(\text{exactly 2 categories}) = n(G \cap W) - 2 + n(W \cap L) - 2 + n(L \cap G) - 2 = 24$ | M1, M1 ft, A1 | Inclusion-exclusion with 3 sets or a 3-set Venn diagram; 63, 2 and 35; 24 as final answer |
## (ii)
| Each row = 20 seats = 10 pairs. There must be (at least) 11 of the people in one row of 10 pairs. So at least one pair with both seats used. | M1, A1 | Using pairs of seats; Using pigeonhole principle |
## (iii)
| e.g. ABC ADE AFG BDF BEG CDG CEF = 7 rounds. $7 + 7 + 7 = 21$ | M1, A1, E1 ft | Listing triples; 7 (rounds/triples), cao; $3 \times$ their 7 |
---50 people are at a TV game show. 21 of the 50 are there to take part in the game show and the others are friends who are in the audience, 22 are women and 20 are from London, 2 are women from London who are there to take part in the game show and 15 are men who are not from London and are friends who are in the audience.
\begin{enumerate}[label=(\roman*)]
\item Deduce how many of the 50 people are in two of the categories 'there to take part in the game show', 'is a woman' and 'is from London', but are not in all three categories. [3]
\end{enumerate}
The 21 people who are there to take part in the game show are moved to the stage where they are seated in two rows of seats with 20 seats in each row. Some of the seats are empty.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show how the pigeonhole principle can be used to show that there must be at least one pair of these 21 people with no empty chair between them. [2]
\end{enumerate}
The 21 people are split into three sets of 7. In each round of the game show, three of the people are chosen. The three people must all be from the same set of 7 but once two people have played in the same round they cannot play together in another round. For example, if A plays with B and C in round 1 then A cannot play with B or with C in any other round.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item By first considering how many different rounds can be formed using the first set of seven people, deduce how many rounds there can be altogether. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete 2018 Q3 [8]}}