a plane is determined by 3 noncollinear pts., how many planes determined by 12 points, no 3 are collinear?

Answer 1

There is a key concept missing in the (otherwise) correct answers above. The points could be co-planar. You could have 12 points, no three of which are collinear (on the same line) but all 12 are in the same plane. For instance the 12 points {(sin(n),cos(n),0)|n=1..12}. That will be a kudo for you if you go to class with that, an "Aha!"

The question should've read: how many planes if there are 12 points, no 3 collinear and no 4 coplanar. Then the answer is 12C3 as indicated in other answers.

Answer 2

The number of planes is determined by the number of combinations you can make from the 12 points choosing them 3 at a time. The number of planes is:

(12*11*10)/3! = (12*11*10)/6 = 220 planes

Answer 3

There is at most one distinct plane for each combination of three points. So how many ways are there to select three points from 12? There are 12 choices for the first point, 11 for the second point, and 10 for the third point. However, the order in which we choose the points doesn't matter, and there are six different ways to choose each combination of three points (thee choices for which points is chosen first and two choices for which opint is chosen second), so that accounting will count every combination 6 times, thus we must divide by 6 to get the final answer. Thus there are at most 12*11*10/6 or 220 distinct planes.

Edit: changed the answer to reflect the fact that the problem did not actually specify that no four points are coplanar (good catch a_math_guy).

Edit 2: although you could say that these points specify exactly 220 planes, not all of which are necessarily distinct.

Answer 4

12C3 = 12*11*10/1/2/3 = 220