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No of triangles

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No of triangles

by srcc25anu » Wed Mar 09, 2011 4:03 pm
How many triangles can be drawn with 17 points, if 5, 7 and 8 of them are in three different lines? The three lines intersect each other in pairs.
1. 680
2. 136
3. 569
4. 171

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by tomada » Thu Mar 10, 2011 4:13 pm
Where did you find that problem?

srcc25anu wrote:How many triangles can be drawn with 17 points, if 5, 7 and 8 of them are in three different lines? The three lines intersect each other in pairs.
1. 680
2. 136
3. 569
4. 171
I'm really old, but I'll never be too old to become more educated.

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by stormier » Thu Mar 10, 2011 5:25 pm
srcc25anu wrote:How many triangles can be drawn with 17 points, if 5, 7 and 8 of them are in three different lines? The three lines intersect each other in pairs.
1. 680
2. 136
3. 569
4. 171
Number of 3 point combinations = 17C3 = 680

Of these there are several 3-point combinations that lie on a straight line and do not form a triangle, so we need to discount them.

Number of 3 point combinations that lie on a straight line = 8C3 + 5C3 + 7C3 = 101

Thus number of unique triangles = 680 - 101 = 579

This is a really tough problem. What's the OA ?

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by srcc25anu » Thu Mar 10, 2011 7:00 pm
Stormier - I too got the same answer 579 using the same methodology that u have applied. the OA is 569 though.
Tomada - And i dont exactly recall the source of this problem .... just happened to come across it on some forum.
stormier wrote:
srcc25anu wrote:How many triangles can be drawn with 17 points, if 5, 7 and 8 of them are in three different lines? The three lines intersect each other in pairs.
1. 680
2. 136
3. 569
4. 171
Number of 3 point combinations = 17C3 = 680

Of these there are several 3-point combinations that lie on a straight line and do not form a triangle, so we need to discount them.

Number of 3 point combinations that lie on a straight line = 8C3 + 5C3 + 7C3 = 101

Thus number of unique triangles = 680 - 101 = 579

This is a really tough problem. What's the OA ?