Problem Solving

hitmis wrote:Why should we not have to also subtract combinations of (x1,y1) = (x2,y2)=(x3,y3) , 9 such combinations, which would also not create a triangle.?

We dont have to. Because these selections are already removed when calculating 9C3.
Evaluate 9C3
9C3=9!*8!*7!/(3!*2!*1!)

Here, 8and7 shows reduced choices. Hence, repeated choices are already removed.

Good question

Good question

Good question

Good question.

9C3-3-3-2 = 76

Answer is 84 Triangles

IMO: 76 is correct.

Because 84-6(3 points in a straight line)=76.

Goddin wrote:IMO: 76 is correct.

Because 84-6(3 points in a straight line)=76.

76 is right ... but we might want to check that arithmetic

The answer would be 76. The total number of combinations would be 9c3= 84, so the answer needs to be something less than this. If you draw the figure and mark all possible points on the x and y axis(it would be a rectangle), you will clearly see that there are 8 lines that would be formed(6 horizontal and vertical and 2 diagonal). If we subtract these 8. We get 84-8=76.

Total 3 points' combinations = 9C3

nCr = n! / ((n-r)! . r!)

9C3 = 9! / 6!.3! = 9.8.7 / 3.2 = 84

Out of 84 combination , 8 will be straight lines

84 - 8 = 76