| Exam Board | OCR |
|---|---|
| Module | Further Discrete (Further Discrete) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Pigeonhole principle applications |
| Difficulty | Standard +0.3 This is a straightforward application of basic counting principles. Part (a) is trivial pigeonhole principle (divide 150 by 7 stacks). Parts (b-d) require systematic counting of digits avoiding 2,3,5, which is routine combinatorial enumeration, and standard inclusion-exclusion with given values. All parts are mechanical applications of standard techniques with no novel insight required, making this easier than average. |
| Spec | 7.01c Pigeonhole principle7.01k Inclusion-exclusion: for two sets7.01l Inclusion-exclusion: extended to more than two sets |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | 150 cards shared equally between 7 (= 6 players and stack) = 21.43 |
| So at least one has 22 or more cards | B1 | |
| [1] | 2.2a | Or 217 = 147 < 150 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | One-digit numbers: 1, 4, 6, 7, 8, 9 are not red = 6 cards |
| Answer | Marks |
|---|---|
| 6 + 42 + 21 = 69 cards with no red digits | M1 |
| A1 | 1.1 |
| 1.1 | Listing/describing 6 one-digit numbers with no red |
| Answer | Marks |
|---|---|
| One-digit numbers: 2, 3, 5 are red = 3 cards | Listing/describing 3 one-digit numbers with a red digit |
| Two-digit numbers: 9 + 21 + 18 = 48 | and making a good attempt at counting the number of |
| RR = 3×3 = 9; RN = 3×7 = 21; NR = 6×3 = 18 | two-digit numbers with at least one red digit |
| Three-digit numbers: 6 + 15 + 9 = 30 | Or appropriate sight of any two of 3, 48, 30 |
| 1RR = 2×3 = 6; 1RN = 2×7 + 1 = 15; 1NR = 3×3 = 9 | Or implied from 81 or from final answer 69 |
| Answer | Marks |
|---|---|
| 150 – 81 = 69 cards with no red digits | Final answer 69 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (c) | Cards in set A are multiples of 2, so units digit is 0, 2, 4, 6 or 8 |
| Answer | Marks |
|---|---|
| 3+24+12 = 39 cards in A with no red digits | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | 4, 6, 8 or 0, 4, 6, 8 as units digit |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (d) | Cards in C with no red digits must have units digit 0, so n(A C) = 9 |
| Answer | Marks |
|---|---|
| = 21 cards | M1* |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | n(A C) = 9 and n(A B C) = 2 seen or implied |
Question 3:
3 | (a) | 150 cards shared equally between 7 (= 6 players and stack) = 21.43
So at least one has 22 or more cards | B1
[1] | 2.2a | Or 217 = 147 < 150
(conclusion may be implied, since given in question)
3 | (b) | One-digit numbers: 1, 4, 6, 7, 8, 9 are not red = 6 cards
Two-digit numbers: tens digit must be 1, 4, 6, 7, 8, 9
and units digit can be any of 0, 1, 4, 6, 7, 8, 9 = 67 = 42 cards
Three-digit numbers: Hundreds digit is 1, tens digit must be 0, 1, 4 and
units digit can be any of 0, 1, 4, 6, 7, 8, 9 = 37 = 21 cards
6 + 42 + 21 = 69 cards with no red digits | M1
A1 | 1.1
1.1 | Listing/describing 6 one-digit numbers with no red
digits and making a good attempt at counting the
number of two-digit numbers with no red digits
Or appropriate sight of any two of 6, 42, 21
Or implied from final answer 69
Final answer 69
Alternative method
One-digit numbers: 2, 3, 5 are red = 3 cards | Listing/describing 3 one-digit numbers with a red digit
Two-digit numbers: 9 + 21 + 18 = 48 | and making a good attempt at counting the number of
RR = 3×3 = 9; RN = 3×7 = 21; NR = 6×3 = 18 | two-digit numbers with at least one red digit
Three-digit numbers: 6 + 15 + 9 = 30 | Or appropriate sight of any two of 3, 48, 30
1RR = 2×3 = 6; 1RN = 2×7 + 1 = 15; 1NR = 3×3 = 9 | Or implied from 81 or from final answer 69
3 + 48 + 30 = 81 cards with at least one red digit
150 – 81 = 69 cards with no red digits | Final answer 69
[2]
3 | (c) | Cards in set A are multiples of 2, so units digit is 0, 2, 4, 6 or 8
One-digit numbers: 4, 6, 8 are not red = 3 cards
Two-digit numbers: tens digit can be any of 1, 4, 6, 7, 8, 9 and units
digit can be any of 0, 4, 6, 8 = 64 = 24 cards
Three-digit numbers: tens digit must be 0, 1, 4 = 34 = 12 cards
3+24+12 = 39 cards in A with no red digits | M1
A1
[2] | 3.1a
1.1 | 4, 6, 8 or 0, 4, 6, 8 as units digit
Trying to count even cards (or odd cards) with no red
digits (or with red digits)
Or implied from final answer 39
Final answer 39
3 | (d) | Cards in C with no red digits must have units digit 0, so n(A C) = 9
60 and 90 are both even, so n(A B C) = 2
69 – (39 + 21 + 9) + ( 12 + 9 + 2) – 2
= 21 cards | M1*
M1dep*
A1
[3] | 1.1
1.1
1.1 | n(A C) = 9 and n(A B C) = 2 seen or implied
Using inclusion-exclusion
Their 69 (from part (b)) – their 39 (from part(c)) – 9
21 from correct working seen3 Six people play a game with 150 cards. Each player has a stack of cards in front of them and the remainder of the cards are in another stack on the table.
\begin{enumerate}[label=(\alph*)]
\item Use the pigeonhole principle to explain why at least one of the stacks must have at least 22 cards in it.
The set of cards is numbered from 1 to 150 .
Each digit '2', '3' and '5', whether as a units digit or a tens digit, is coloured red.\\
So, for example
\begin{itemize}
\item the card numbered 25 has two red digits,
\item the card numbered 26 has one red digit,
\item the card numbered 148 has no red digits.
\item By considering the cards with one digit, two digits and three digits, or otherwise, determine how many cards have no red digits.
\end{itemize}
The cards are put into a Venn diagram with three intersecting sets:\\
$\mathrm { A } = \{$ cards with a number that is a multiple of $2 \}$\\
$\mathrm { B } = \{$ cards with a number that is a multiple of $3 \}$\\
$\mathrm { C } = \{$ cards with a number that is a multiple of $5 \}$
For example
\begin{itemize}
\item the card numbered 2 is in set A ,
\item the card numbered 15 is in sets B and C ,
\item the card numbered 23 is in none of the sets.\\
\includegraphics[max width=\textwidth, alt={}, center]{133395d2-5020-4054-a229-70168f1d0f95-4_588_1150_1667_246}
\item By considering the cards with one digit, two digits and three digits, or otherwise, determine how many cards in set A have no red digits.
\item Given that, for the cards with no red digits, $n ( B ) = 21 , n ( C ) = 9$ and $n ( A \cap B ) = 12$, use the inclusion-exclusion principle to determine how many of the cards with no red digits are in none of the sets A, B or C.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete 2021 Q3 [8]}}