| Exam Board | OCR |
|---|---|
| Module | Further Discrete (Further Discrete) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Correct ordering probability |
| Difficulty | Standard +0.8 This is a Further Maths question on derangements requiring multiple conceptual steps: computing specific restricted arrangements, proving a derangement formula exists, and deriving a recurrence relation. Part (iv) requires genuine combinatorial insight beyond routine calculation, placing it moderately above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities7.01e Permutations: ordered subsets of r from n elements7.01f Combinations: unordered subsets of r from n elements7.01g Arrangements in a line: with repetition and restriction7.01m Derangements: enumeration by ad hoc methods |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | 5! or 5P (cid:32)120 |
| 5 | B1 | |
| [1] | 1.1 | |
| 3 | (ii) | 5C (cid:32)10 ways to choose the three letters that |
| Answer | Marks |
|---|---|
| exactly 3 letters in the correct envelopes. | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Any valid attempt |
| Answer | Marks |
|---|---|
| 10 from valid reasoning | A list of the ten possibilities |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (iii) | (a) |
| Answer | Marks |
|---|---|
| by A and then B (so that B is not second). | B1 |
| [1] | 1.1 |
| Answer | Marks |
|---|---|
| c | m |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (iii) | (b) |
| Answer | Marks |
|---|---|
| in the correct envelopes. | e |
| Answer | Marks | Guidance |
|---|---|---|
| [1] | 2.2a | 20 from valid reasoning |
| 3 | (iv) | There are n(cid:16)1 choices for the first position. |
| Answer | Marks |
|---|---|
| Hence the result given. | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| Answer | Marks |
|---|---|
| c | Or any equivalent partial argument |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (v) | D(cid:11)4(cid:12)(cid:32)3(cid:11)2(cid:14)1(cid:12)(cid:32)9 p |
| Answer | Marks |
|---|---|
| be in the wrong envelopes | e |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 3.2a | OR 1 way with all five correct, 10 ways |
| Answer | Marks |
|---|---|
| 120(cid:16)(cid:11)45(cid:14)20(cid:14)10(cid:14)1(cid:12)(cid:32)120(cid:16)76(cid:32)44 | OR |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | –1 | 2 |
Question 3:
3 | (i) | 5! or 5P (cid:32)120
5 | B1
[1] | 1.1
3 | (ii) | 5C (cid:32)10 ways to choose the three letters that
3
will be put in the correct envelopes.
The other two letters must be swapped over, so
there is only one way to put them in the wrong
envelopes.
So there are 10(cid:117)1(cid:32)10 arrangements with
exactly 3 letters in the correct envelopes. | M1
E1
[2] | 1.1
1.1 | Any valid attempt
n
e
10 from valid reasoning | A list of the ten possibilities
(cid:159)M1A0 unless supported
with reasoning to show that
there are no other possibilities
3 | (iii) | (a) | The first symbol cannot be A. If it is B then it
must be followed by C and then A (so that C is
not last) and if it is C then it must be followed
by A and then B (so that B is not second). | B1
[1] | 1.1
i
c | m
Or list the six permutations and identify
the two derangements (but not just
writing down the two derangements with
no explanation for why there are no
others)
3 | (iii) | (b) | 5C (cid:32)10 ways to choose the two lett p ers that
2
will be put into the correct envelopes and 2
S
ways to arrange the other three letters so that
none are in the correct envelopes. So there are
10(cid:117)2(cid:32)20 arrangements with exactly 2 letters
in the correct envelopes. | e
E1
[1] | 2.2a | 20 from valid reasoning
3 | (iv) | There are n(cid:16)1 choices for the first position.
Then the symbol that was in the first position
either goes in the position corresponding to the
symbol that is now in the first position or not.
In the first case we are left with n(cid:16)2 symbols
with the corresponding n(cid:16)2 positions, which
can be done D(cid:11)n(cid:16)2(cid:12)ways.
In the second case we can rename the first
symbol with the name of the second symbol
and then we have n(cid:16)1 symbols with n(cid:16)1
corresponding positions, which can be done
D(cid:11)n(cid:16)1(cid:12)ways.
Hence the result given. | M1
E1
[2] | 3.1a
2.1
i
c | Or any equivalent partial argument
Achieving the given result
D(cid:11)n(cid:12)(cid:32)n(cid:11)n(cid:16)1(cid:12)(cid:117) (cid:11)D(cid:11)n(cid:16)1(cid:12)(cid:14)D(cid:11)n(cid:16)2(cid:12)(cid:12)
e
m
3 | (v) | D(cid:11)4(cid:12)(cid:32)3(cid:11)2(cid:14)1(cid:12)(cid:32)9 p
D(cid:11)5(cid:12)(cid:32)4(cid:11)9(cid:14)2(cid:12)(cid:32)44
S
There are 44 ways in which all five letters can
be in the wrong envelopes | e
M1
A1
[2] | 3.1a
3.2a | OR 1 way with all five correct, 10 ways
with three correct, 20 ways with two
correct and 5(cid:117)9(cid:32)45 ways with one
correct
120(cid:16)(cid:11)45(cid:14)20(cid:14)10(cid:14)1(cid:12)(cid:32)120(cid:16)76(cid:32)44 | OR
M1 use inclusion-exclusion:
5! 5! 5! 5! 5!
5!(cid:16) (cid:14) (cid:16) (cid:14) (cid:16)
1! 2! 3! 4! 5!
(cid:32)120(cid:16)120(cid:14)60(cid:16)20(cid:14)5(cid:16)1
A1 (cid:32)44
3 | –1 | 23 Bob has been given a pile of five letters addressed to five different people. He has also been given a pile of five envelopes addressed to the same five people. Bob puts one letter in each envelope at random.
\begin{enumerate}[label=(\roman*)]
\item How many different ways are there to pair the letters with the envelopes?
\item Find the number of arrangements with exactly three letters in the correct envelopes.
\item (a) Show that there are two derangements of the three symbols A , B and C .\\
(b) Hence find the number of arrangements with exactly two letters in the correct envelopes.
Let $\mathrm { D } _ { n }$ represent the number of derangements of $n$ symbols.
\item Explain why $\mathrm { D } _ { n } = ( n - 1 ) \times \left( \mathrm { D } _ { n - 1 } + \mathrm { D } _ { n - 2 } \right)$.
\item Find the number of ways in which all five letters are in the wrong envelopes.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete Q3 [9]}}