Consider a regular octagon.  How many triangles can be formed whose vertices are the vertices of the octagon?
I need to count the number of ways to choose three vertices out of eight, without repetition or regard to order. This is a combination problem, and I can use the formula for n choose r, which is nCr = n!/(r!(n-r)!), where ! means factorial. In this case, n is 8 and r is 3, so I get 8C3 = 8!/(3!(8-3)!) = (8*7*6*5*4*3*2*1)/((3*2*1)(5*4*3*2*1)) = (8*7*6)/(3*2*1) = 56. 