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anonymous
 3 years ago
Help with mathematical combinations please.
Points A, B, C, D, E, F, G, H, and I are arranged in two rows such that we have A  B  C  D  E and
F  G  H  I
How many triangles can be formed having vertices chosen from these points?
How many of these triangles have the point A as one of their vertices?
anonymous
 3 years ago
Help with mathematical combinations please. Points A, B, C, D, E, F, G, H, and I are arranged in two rows such that we have A  B  C  D  E and F  G  H  I How many triangles can be formed having vertices chosen from these points? How many of these triangles have the point A as one of their vertices?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry it is A  B  C  D  E F  G  H  I

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Awesome thank you for visiting unklerhaukus!!!

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1360476356312:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes they are to act as the vertices of triangles.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1so either two points are from the top line (and one from the bottom line) or one point is from the top line (and two from the bottem line)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm i see, let me try that, i wasnt sure how to interpret it.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1because if all three points were on the same line , you couldn't make a triangle

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes that appears to work, 5C1 x 4C2 and 5C2 x 4C1 which adds to 70

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Though can I ask how you determine how to do these questions? I havent really got a problem solving type of mind, so i find it difficult to think laterally.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[^nC_k = \frac{n!}{k!(nk)!}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No i understand the general formula of the combination theory. I just often have difficulty understanding how I am meant to approach the question, what guidelines I need to follow.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Once you had explained those guidelines i knew exactly what to do, i just dont know how to determine the parameters to follow.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1i like to start with a picture usually

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are there any other tricks you can use to figure out whats going on? or is it just trying to pull apart the question, and making assumptions to rules.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1im not sure how to answer that,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah well, im sure i'll just have to pick it up as I go. You've helped enough already. Thank you very much for the assistance. I wish you a good afternoon. Thank you for the help mate, I appreciate the effort :) Have a good one mate, and stay well!

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1sometimes rewriting the question into a simpler form can help
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