The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is
- 56
- 16
- 24
- 48
Answer
(2) 16
Solution
Each vertex of such a triangle must be non-adjacent to the other two.
Fix one vertex (8C1 = 8 ways). Its two neighbours are barred, so the other two vertices come from the remaining 8 - 3 = 5 vertices, which lie in a row and must not be adjacent to each other: 5 - 1C2 = 4C2 = 6 ways (gap method).
Since every triangle is counted once from each of its 3 vertices,
number = 8C1 4C23 = 8 63 = 16
(For an n-gon this count is n n - 4C23.)
(1): 56 is 8C3, all triangles, so it ignores the condition; (4): 48 forgets to divide by 3; (3): 24 matches no correct count. So only (2) is right.
Answer: (2)