| Exam Board | OCR |
|---|---|
| Module | Further Discrete (Further Discrete) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Challenging +1.2 This is a systematic multi-part combinations question requiring careful case-by-case enumeration in parts (c)-(e). While part (a) is routine C(7,4), parts (b)-(c) require organized counting with constraints, and parts (d)-(e) involve permutations with the subtlety of handling repeated scores. The constraint checking (total ≤20) and non-decreasing arrangement counting require methodical work but no deep insight—slightly above average difficulty for Further Maths. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | 35 |
| [1] | 1.1 | 7C 4 or 7×6×5×4 o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | 3 men, 1 woman = 5C 2C = 10 2 = 20 |
| Answer | Marks |
|---|---|
| 20 + 5 = 25 | M1 |
| A1 | 2.1 |
| 1.1 | Attempting either case (some appropriate working seen) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | Or equivalent |
| 35 – 10 = 25 | A1 FT | 25 or (their 35) – 10 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (c) | 3 4 4 with 6, 7, 8 or 9 |
| Answer | Marks |
|---|---|
| Total = 6 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.1 | Identifying at least 4 valid cases |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (d) | Each team can be arranged in 4! = 24 ways |
| 6 24 = 144 | B1 FT | |
| [1] | 1.1 | 144 or (their 6) 24 |
| 3 | (e) | 3 4 4 with 6, 7, 8 or 9 = 2 4 = 8 |
| Answer | Marks |
|---|---|
| Total = 10 | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 1.1 | Recognising that the two 4’s are distinguishable |
Question 3:
3 | (a) | 35 | B1
[1] | 1.1 | 7C 4 or 7×6×5×4 o.e.
4!
3 | (b) | 3 men, 1 woman = 5C 2C = 10 2 = 20
3 1
4 men = 5C = 5
4
20 + 5 = 25 | M1
A1 | 2.1
1.1 | Attempting either case (some appropriate working seen)
25
Alternative method
2 men, 2 women = 5C 2C = 10 1 = 10
2 2 | M1 | Or equivalent
35 – 10 = 25 | A1 FT | 25 or (their 35) – 10
[2]
3 | (c) | 3 4 4 with 6, 7, 8 or 9
3 and one of the 4’s with 6, 7
4 ways for 3, 4, 4, with 6, 7, 8 or 9
2 ways for 3 one of the 4’s with 6, 7
Total = 6 | M1
A1
[2] | 3.1b
2.1 | Identifying at least 4 valid cases
{3, 4, 4′, X} with X = 6, 7, 8 or 9
{3, 4, 6, 7}, (3, 4′, 6, 7}
6 from valid working
3 | (d) | Each team can be arranged in 4! = 24 ways
6 24 = 144 | B1 FT
[1] | 1.1 | 144 or (their 6) 24
3 | (e) | 3 4 4 with 6, 7, 8 or 9 = 2 4 = 8
3 4 6 7 with either 4 = 2
Total = 10 | M1
A1 FT
[2] | 2.2a
1.1 | Recognising that the two 4’s are distinguishable
(some appropriate working seen, e.g. allow 2 (their 6)
or implied from answer 10 if (c) was correct)
10
or follow through their (incomplete) list of valid cases with no
extras (= 2 (their 4) + (their 2))3 A para relay team of 4 swimmers needs to be chosen from a group of 7 swimmers.
\begin{enumerate}[label=(\alph*)]
\item How many ways are there to choose 4 swimmers from 7?
There are no restrictions on how many men and how many women are in the team. The group of 7 swimmers consists of 5 men and 2 women.
\item How many ways are there to choose a team with more men than women?
The physical impairment of each swimmer is given a score.\\
The scores for the swimmers are\\
$\begin{array} { l l l l l l l } 3 & 4 & 4 & 6 & 7 & 8 & 9 \end{array}$
The total score for the team must be 20 or less.
\item How many different valid teams are possible?
The order of the swimmers in the team is now taken into consideration.
\item In total, how many different arrangements are there of valid teams?
\item In how many of these valid teams are the scores of the swimmers in increasing order? For example, 3, 4, 4, 8 but not 4, 3, 4, 8 .
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete 2022 Q3 [8]}}