| Exam Board | OCR |
|---|---|
| Module | Further Discrete AS (Further Discrete AS) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Standard +0.3 This is a straightforward permutations problem with repeated elements. Part (a) requires basic counting with distinct digits, part (b) needs careful case analysis with one pair, and part (c) requires systematic enumeration of all cases (all different, one pair, two pairs, three same, all same). While multi-part and requiring organization, the techniques are standard for Further Maths permutations with no novel insight needed. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 1 | 2 |
| 3 | 2 | 2 |
| 3 | (a) | 9 8 7 6 or 9P |
| Answer | Marks |
|---|---|
| = 3024 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Appropriate working seen or explained |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | (9 8 7) 6 |
| = 3024 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Sight of calculation 504 N or 252 N o.e., for positive integer N |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (c) | 94 |
| = 6561 | M1 | |
| A1 | 2.1 | |
| 1.1 | soi |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3024 + 3024 + 216 + 288 + 9 = 6561 | A1 | 6561 |
Question 3:
3 | 1 | 2 | 1 | 2
3 | 2 | 2 | 1 | 0
3 | (a) | 9 8 7 6 or 9P
4
= 3024 | M1
A1
[2] | 1.1
1.1 | Appropriate working seen or explained
cao
SC B1 3024 seen with no relevant working
3 | (b) | (9 8 7) 6
= 3024 | M1
A1
[2] | 1.1
1.1 | Sight of calculation 504 N or 252 N o.e., for positive integer N
OR a calculation leading to final answer 3024, 6048, 1512 or 504
cao
SC B1 3024 seen with no relevant working
3 | (c) | 94
= 6561 | M1
A1 | 2.1
1.1 | soi
cao
Alternative solution
aabb 9C 6 = 216
2 | M1 | M1 | At least two of 216, 288, 9 | At least two of 216, 288, 9
aaab 9P 4 = 288
2
aaaa 9
3024 + 3024 + 216 + 288 + 9 = 6561 | A1 | 6561
[2]3 Heidi has a pack of cards.\\
Each card has a single digit on one side and is blank on the other side.\\
Each of the digits from 1 to 9 appears on exactly four cards.\\
Apart from the numerical values of the digits, the cards are indistinguishable from each other.\\
Heidi draws four cards from the pack, at random and without replacement. She places the four cards in a row to make a four-digit number.
Determine how many different four-digit numbers Heidi could have made in each of the following cases.
\begin{enumerate}[label=(\alph*)]
\item The four digits are all different.
\item Two of the digits are the same and the other two digits are different.
\item There is no restriction on whether any of the digits are the same or not.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete AS 2024 Q3 [6]}}