If one of two parallel lines has 7 points on it and the other has 8, how many unique triangles can be drawn using the points?
A) 168
B) 196
C) 316
D) 364
E) 455
OA: D
how?
how many unique triangles?
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- thephoenix
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to make a triangle we need 3 points . here two case r possible 1 point frm line of 7 points and 2 point frm line of 8 points
or
1 point from 8 and 2 frm 7
which is as:-
8c2*7c1+7c2*8c1=364
or
1 point from 8 and 2 frm 7
which is as:-
8c2*7c1+7c2*8c1=364
- ajith
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If none of the points were collinear the number of triangles possible = 15C3 = 15*14*13/6 = 455rahul.s wrote:If one of two parallel lines has 7 points on it and the other has 8, how many unique triangles can be drawn using the points?
A) 168
B) 196
C) 316
D) 364
E) 455
OA: D
how?
Triangles not possible because of 8 collinear points on a straight line = 8C3 = 56
Triangles not possible because of 8 collinear points on a straight line = 7C3 = 35
Total possible triangles = 455 -46-35 =364
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I like both of the above solutions, and think they are probably the most natural approaches to take here. There is at least one other way to solve the problem:
-we must choose one point from the first line (7 choices)
-we must choose one point from the second line (8 choices)
-we must choose one of the remaining points (13 choices)
-the order of the two points chosen from the same line doesn't matter (divide by 2!)
so the answer is (7*8*13)/2 = 364
You might notice that this problem is just a geometric version of this counting problem:
"A company employs 7 men and 8 women. How many different three-person committees could be formed which do not consist entirely of men or entirely of women?"
This counting problem format, where we must choose fixed numbers of things from two different groups, is very common on the GMAT.
-we must choose one point from the first line (7 choices)
-we must choose one point from the second line (8 choices)
-we must choose one of the remaining points (13 choices)
-the order of the two points chosen from the same line doesn't matter (divide by 2!)
so the answer is (7*8*13)/2 = 364
You might notice that this problem is just a geometric version of this counting problem:
"A company employs 7 men and 8 women. How many different three-person committees could be formed which do not consist entirely of men or entirely of women?"
This counting problem format, where we must choose fixed numbers of things from two different groups, is very common on the GMAT.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
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- money9111
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can someone tell me the level of question this would be?
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There's an identical problem (but with smaller numbers of points) in GMATFocus, and you only encounter it if you're doing quite well, so the difficulty level is certainly well above average.money9111 wrote:can someone tell me the level of question this would be?
Incidentally, I'm curious where the question in the post above comes from, since it is exactly the same as a real GMAT question but with slightly larger numbers.
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- money9111
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ok so i'll worry about this level question, when I in fact get up to that level... otherwise I'll never see an example like this...
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"A company employs 7 men and 8 women. How many different three-person committees could be formed which do not consist entirely of men or entirely of women?"
Just curious.....the solution to above problem should be -
15C3 - 7C3 - 8C3
Can someone please confirm?
Just curious.....the solution to above problem should be -
15C3 - 7C3 - 8C3
Can someone please confirm?
- ajith
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Answer to that problem is 15C3 - 7C3 - 8C3.teal wrote:"A company employs 7 men and 8 women. How many different three-person committees could be formed which do not consist entirely of men or entirely of women?"
Just curious.....the solution to above problem should be -
15C3 - 7C3 - 8C3
Can someone please confirm?
Confirmed
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