OCR Further Discrete 2023 June — Question 4 10 marks

Exam BoardOCR
ModuleFurther Discrete (Further Discrete)
Year2023
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCombinations & Selection
TypeMulti-stage selection problems
DifficultyChallenging +1.2 This is a multi-part combinations question requiring systematic application of selection formulas and inclusion-exclusion principle. Parts (a)-(b) are standard Further Maths counting problems, while part (c) requires careful case analysis with inclusion-exclusion, and part (d) is a straightforward pigeonhole principle application. The question is more computational than conceptually challenging, though the inclusion-exclusion in part (c) elevates it slightly above average Further Maths difficulty.
Spec5.01a Permutations and combinations: evaluate probabilities7.01c Pigeonhole principle7.01e Permutations: ordered subsets of r from n elements7.01f Combinations: unordered subsets of r from n elements
4 The first 20 consecutive positive integers include the 8 prime numbers \(2,3,5,7,11,13,17\) and 19. Emma randomly chooses 5 distinct numbers from the first 20 consecutive positive integers. The order in which Emma chooses the numbers does not matter.
  1. Calculate the number of possibilities in which Emma's 5 numbers include exactly 2 prime numbers and 3 non-prime numbers.
  2. Calculate the number of possibilities in which Emma's 5 numbers include at least 2 prime numbers. The pairs \(\{ 3,13 \}\) and \(\{ 7,17 \}\) each consist of numbers with a difference of exactly 10 .
  3. Calculate the number of possibilities in which Emma's 5 numbers include at least one pair of prime numbers in which the difference between them is exactly 10 . A new set of 20 consecutive positive integers, each with at least two digits, is chosen. This set of 20 numbers contains 5 prime numbers.
  4. Use the pigeonhole principle to show that there is at least one pair of these prime numbers for which the difference between them is exactly 10 .
Question 4:
AnswerMarks Guidance
4(a) 2 primes from 8 and 3 non-primes from 12
8C  12C = 28  220
2 3
AnswerMarks
= 6160M1
A1
AnswerMarks
[2]1.1
1.18C  12C or 28  220 (o.e.) seen (isw)
2 3
cao
AnswerMarks Guidance
4(b) 6160 + (8C  12C ) + (8C  12C ) + 8C
3 2 4 1 5
= 6160 + (56  66) + (70  12) + 56
= 6160 + 3696 + 840 + 56
AnswerMarks
= 10752M1*
M1 ft
dep*
AnswerMarks
A1 ft3.1a
1.1
AnswerMarks
1.1Any of (8C  12C ), (8C  12C ), 8C (o.e.) seen
3 2 4 1 5
Their attempt at 6160 + 3696 + 840 + 56 (soi)
FT their 6160
10752 or 4592 + (their) 6160 evaluated
Alternative method
20C – (8C  12C ) – 12C
AnswerMarks Guidance
5 1 4 5M1* Any of 20C , (8C  12C ), 12C (o.e.) in any form
5 1 4 5
= 15504 – (8  495) – 792
AnswerMarks Guidance
= 15504 – 3960 – 792M1
dep *Their attempt at 15504 – 3960 – 792 (soi)
= 10752A1 10752 cao
[3]
AnswerMarks Guidance
4(c) {3, 13} and 3 others (but not both 7 and 17)
or {7, 17}and 3 others (but not both 3 and
13) or {3, 7, 13, 17}and 1 other
= 18C + 18C – 16C
3 3 1
= 816 + 816 – 16
AnswerMarks
= 1616M1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1Any of 18C , 16C , 17C , (= 680) or 16C (= 120) o.e. seen
3 1 3 2
Or any of 816, 800, 680, 120 or 16 seen as values to be + or –
or implied from final answer
Their attempt at 816 + 816 – 16 or 800 + 800 + 16 or 816 + 800
Allow any two of 816, 816, 800, 800 with  16 added
Or 680 + 680 + 120 + 120 + 16 (o.e.)
May be implied from final answer 1600, 1616, 1632, provided
first M mark has been awarded
cao
AnswerMarks Guidance
4(d) Units digit  {1, 3, 7, 9}
5 > 4 so at least two primes with the same
AnswerMarks
units digit, hence resultM1
A1
AnswerMarks
[2]2.5
2.4Identify these four pigeonholes
e.g. units digit is not even and is not 5
Using pigeonhole principle to draw conclusion
Question 4:
4 | (a) | 2 primes from 8 and 3 non-primes from 12
8C  12C = 28  220
2 3
= 6160 | M1
A1
[2] | 1.1
1.1 | 8C  12C or 28  220 (o.e.) seen (isw)
2 3
cao
4 | (b) | 6160 + (8C  12C ) + (8C  12C ) + 8C
3 2 4 1 5
= 6160 + (56  66) + (70  12) + 56
= 6160 + 3696 + 840 + 56
= 10752 | M1*
M1 ft
dep*
A1 ft | 3.1a
1.1
1.1 | Any of (8C  12C ), (8C  12C ), 8C (o.e.) seen
3 2 4 1 5
Their attempt at 6160 + 3696 + 840 + 56 (soi)
FT their 6160
10752 or 4592 + (their) 6160 evaluated
Alternative method
20C – (8C  12C ) – 12C
5 1 4 5 | M1* | Any of 20C , (8C  12C ), 12C (o.e.) in any form
5 1 4 5
= 15504 – (8  495) – 792
= 15504 – 3960 – 792 | M1
dep * | Their attempt at 15504 – 3960 – 792 (soi)
= 10752 | A1 | 10752 cao
[3]
4 | (c) | {3, 13} and 3 others (but not both 7 and 17)
or {7, 17}and 3 others (but not both 3 and
13) or {3, 7, 13, 17}and 1 other
= 18C + 18C – 16C
3 3 1
= 816 + 816 – 16
= 1616 | M1
M1
A1
[3] | 3.1a
1.1
1.1 | Any of 18C , 16C , 17C , (= 680) or 16C (= 120) o.e. seen
3 1 3 2
Or any of 816, 800, 680, 120 or 16 seen as values to be + or –
or implied from final answer
Their attempt at 816 + 816 – 16 or 800 + 800 + 16 or 816 + 800
Allow any two of 816, 816, 800, 800 with  16 added
Or 680 + 680 + 120 + 120 + 16 (o.e.)
May be implied from final answer 1600, 1616, 1632, provided
first M mark has been awarded
cao
4 | (d) | Units digit  {1, 3, 7, 9}
5 > 4 so at least two primes with the same
units digit, hence result | M1
A1
[2] | 2.5
2.4 | Identify these four pigeonholes
e.g. units digit is not even and is not 5
Using pigeonhole principle to draw conclusion
4 The first 20 consecutive positive integers include the 8 prime numbers $2,3,5,7,11,13,17$ and 19.

Emma randomly chooses 5 distinct numbers from the first 20 consecutive positive integers. The order in which Emma chooses the numbers does not matter.
\begin{enumerate}[label=(\alph*)]
\item Calculate the number of possibilities in which Emma's 5 numbers include exactly 2 prime numbers and 3 non-prime numbers.
\item Calculate the number of possibilities in which Emma's 5 numbers include at least 2 prime numbers.

The pairs $\{ 3,13 \}$ and $\{ 7,17 \}$ each consist of numbers with a difference of exactly 10 .
\item Calculate the number of possibilities in which Emma's 5 numbers include at least one pair of prime numbers in which the difference between them is exactly 10 .

A new set of 20 consecutive positive integers, each with at least two digits, is chosen. This set of 20 numbers contains 5 prime numbers.
\item Use the pigeonhole principle to show that there is at least one pair of these prime numbers for which the difference between them is exactly 10 .
\end{enumerate}

\hfill \mbox{\textit{OCR Further Discrete 2023 Q4 [10]}}
This paper (7 questions)
View full paper
Q1 7 Q2 8 Q3 12 Q4 10 Q5 12 Q6 14 Q7 12