2022 HSC Mathematics Extension 1
Question 7:
Combinatorics
The diagram shows triangle with points chosen on each of the sides. On side , 3 points are chosen. On side , 4 points are chosen. On side , 5 points are chosen.
How many triangles can be formed using the chosen points as vertices?
How to recognise this question
Question type: combinatorics
- The command and notation point to combinatorics.
- The requested response is a multiple choice.
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
- There are chosen points in total.
- Any 3 of the 12 points determine a triangle unless the 3 are collinear on the same side.
- Total ways to choose 3 points: .
- Collinear exclusions: on , on , on .
- Number of triangles: .
- Answer: C. (NESA multiple-choice key: C.)
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 1 total mark 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
Points: 2 on , 3 on , 4 on . How many triangles can be formed from the chosen points?
Question 2
Three points on each side of a triangle (9 points total). Number of triangles?
Question 3
If 5 points are in general position (no 3 collinear), equals