2023 HSC Mathematics Extension 1
Question 12(d):
Combinatorics
It is known that for all integers such that . (Do NOT prove this.)
Find ONE possible set of values for and such that
How to recognise this question
Question type: combinatorics
- The command and notation point to combinatorics.
- The requested response is a worked solution.
How to handle it: count a simple unrestricted set first, then add or subtract the arrangements that break the condition.
Watch out: Do not select a formula until its domain, interval, sign and units match the question.
Step-by-step answer
- Apply Pascal to the first two terms: .
- So LHS .
- Use symmetry: .
- Hence LHS by Pascal again.
- One solution: , . (Also works by symmetry: .)
- NESA: 2 marks correct; 1 for combining the first two terms with the given identity.
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 2 total marks for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.
Try your hand at it again by answering these questions
Question 1
Find such that .
Question 2
Rewrite using symmetry.
Question 3
Find one pair such that .