Difference between revisions of "2019 Mock AMC 10B Problems/Problem 14"
Rollover2020 (talk | contribs) (Created page with "There are five points, and to make a triangle we need 3, so the number of ways is just <math>\binom(5 3).</math> This is option choice (A). - Armpit Splash 💦") |
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| − | There are five points, and to make a triangle we need 3, so the number of ways is just <math>\ | + | Solution 1 |
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| + | There are five points, and to make a triangle we need 3, so the number of ways is just <math>\binom5 3.</math> This is option choice (A). | ||
- Armpit Splash 💦 | - Armpit Splash 💦 | ||
| + | - Edits by Continuous_Pi | ||
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| + | Note: This solution is wrong because we must consider when a subset of the points are collinear, creating degenerate triangles. | ||
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| + | Note 2: It doesn't matter the problem is just bad and should say that no three of the points are collinear or something because if there were degenerate triangles the answer would be less than 10, but that's not an answer choice so it doesnt matter - lucaswujc | ||
Latest revision as of 16:34, 2 November 2024
Solution 1
There are five points, and to make a triangle we need 3, so the number of ways is just 
This is option choice (A).
- Armpit Splash 💦
- Edits by Continuous_Pi
Note: This solution is wrong because we must consider when a subset of the points are collinear, creating degenerate triangles.
Note 2: It doesn't matter the problem is just bad and should say that no three of the points are collinear or something because if there were degenerate triangles the answer would be less than 10, but that's not an answer choice so it doesnt matter - lucaswujc
