fin with triangles: solution
 

Given fourteen points draw equidistant to each other in the shape of a pyramid missing the top piece (5 bottom row, 4 the row above, 3 the row above that and finally two on the row above that which is the top), determine the number of triangles that can be drawn formed with the vertices at these points.
the picture should look like Pascals Triangle without the top point . I don't know how to draw the image of the dots properly on this website. Imagine 14 dots arranged in a pyramid without a top piece. Please give a brief explanation. good luck, have fun!

-Brent

hint: the answer is a prime number and is not 329, this is a statistics problem, the triangles do not have to be equilateral

solution:
find the total number of groups of three the figure can produce by using combinations (14C3)= 364. Now, subtract the groups of three that are colinear (which are obviously not triangles) also using combinations. 364-3-10-20= 331

331 triangles total

Re: fin with triangles: solution
 

does anyone disagree?

Re: fin with triangles: solution
 

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does anyone disagree?

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You probably missed the 2 colinear sets that I had missed originally. If you enumerate row-wise and going down from 1 to 14, you will find that (3, 8, 14) and (5, 7, 10) are both colinear sets.

I am pretty sure BruceZ is right. After Bruce posted his solution, I went back to take a closer look at mine, and my final corrected solution lines up with his.

329 it is.

Your original thread: Original Thread

Edited to add the specific colinear sets that were likely missing from the OP's solution.

-RMJ

Re: fin with triangles: solution
 

Yeah, I went back at it and realized that the person who gave me the problem is in fact wrong. The correct answer is 329 triangles