2025 HSC Mathematics Extension 1
Question 13(e):
Combinatorics
(i) Starting from Pascal’s relation, show .
(ii) Hence, or otherwise, prove .
How to recognise this question
Question type: combinatorics
- The command and notation point to combinatorics.
- The word “hence” means the later part is intended to reuse the earlier result.
How to handle it: count a simple unrestricted set first, then add or subtract the arrangements that break the condition.
Watch out: Do not restart the second part with unrelated numbers or discard the result proved in the first part.
Step-by-step answer
- Rearrange Pascal’s identity and substitute , to obtain the difference identity in (i).
- Apply that identity to every term in (ii); adjacent binomial coefficients cancel.
- Only the first negative and final positive terms remain.
- The sum simplifies to .
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 3 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
(i) Derive from Pascal’s identity.
(ii) Hence evaluate .
Question 2
(i) Use Pascal’s identity to show .
(ii) Hence evaluate .
Question 3
(i) Show .
(ii) Hence evaluate .