Given that there are 16 dots equally spaced in an array, any unique combination of E does chosen can be a valid triangle with positive area UNLESS those 3 points chosen are COLINEAR
to find the number of valid triangles:
(total possible unique combinations of 3 dots chosen out of the 16 total)
-
(Unfavorable Outcomes in which the 3 dots chosen all lie on the same line)
(16 c 3) = 16! / (3! 13!) = 560 unique combinations can be made of 3 dots
Out of these 560 groupings, how many groupings involve 3 dots located on the same Line
Case 1: vertical rows
For any one vertical row, there are 4 dots. Any 3 unique dots chosen from those 4 dots will lie on a straight line and will NOT form a valid triangle. How many ways can we have unique groups of 3 dots chosen out of 4 in a row?
(4 c 3) = 4 ways
Since there are 4 rows: (4 rows) * (4 combinations per row that do not form a triangle) = 16 invalid
Case 2: Same logic as case 1.
There will be another 16 invalid groupings.
Case 3: 3 dots that lie on Diagonally Upwards Sloping Lines
There are 2 Diagonally upward slopping lines that sandwich the Center Diagonal.
Each of those lines has only 3 dots. Each will be an invalid case ———> 2 invalid
The upward sloping diagonal of the 4 by 4 array contains 4 dots. Any 3 dots chosen from these 4 dots that lie on a diagonal straight line will NOT form a triangle.
(4 c 3) = 4 invalid
6 invalid cases
Case 4: downward sloping diagonal lines
Symmetrically, the same logic will apply as applied in case 3.
Another 6 invalid cases that must be removed.
Total number of possible triangles = (560) - (16) - (16) - (6) - (6) =
(560) - (44) =
516 possible triangles can be made
Answer (A)
Posted from my mobile device