How many straight lines can be formed using 10 points , 4 of which are collinear ?
A) 20
B) 30
C) 40
D) 50
E) 60
OA
C
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How many straight lines can be formed using 10 points
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gmat_guy666
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That's a very vague (ambiguous) question.gmat_guy666 wrote:How many straight lines can be formed using 10 points , 4 of which are collinear ?
A) 20
B) 30
C) 40
D) 50
E) 60
Do the lines need to pass through any of the points? Is passing through 1 point enough? 2 points? 3 points?
Given that the correct answer is C, I assume the question should be worded as follows:
If a line is defined two points, how many different lines can be formed with 10 points, 4 of which are collinear ?
NOTE: IF none of the points were collinear (all in a row), then the number of lines = 10C2 (10 points, choose 2 of them).
10C2 = 45, so if we IGNORE the fact that some points are collinear, the answer is 45.
HOWEVER, we need to subtract the lines that can be created using 2 of the 4 collinear points, since these lines will all be the SAME.
In how many ways can we select 2 points from the 4 collinear points?
We can do this in 4C2 ways (= 6 ways)
In other words, all 6 possible lines are the SAME line.
So, we have counted 5 extra lines.
45 - 5 = 40
Answer: C
Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2 and 4C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Cheers,
Brent
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GMATGuruNY
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A better wording might be as follows:
Let the 4 collinear points be A, B, C and D.
From these 4 points, the number of ways to choose 2 to form a line = 4C2 = (4*3)/(2*1) = 6.
But these 6 pairs all constitute the SAME LINE -- line ABCD.
Since these 6 pairs constitute ONLY 1 LINE, we must subtract 5 of these 6 pairs from our total of 45:
45-5 = 40.
The correct answer is C.
From 10 points, the number of ways to choose 2 to form a line = 10C2 = (10*9)/(2*1) = 45.On the coordinate plane are 10 points, 4 of which are collinear. How many distinct straight lines can be formed by joining any 2 of these points?
A) 20
B) 30
C) 40
D) 50
E) 60
OA
C
Let the 4 collinear points be A, B, C and D.
From these 4 points, the number of ways to choose 2 to form a line = 4C2 = (4*3)/(2*1) = 6.
But these 6 pairs all constitute the SAME LINE -- line ABCD.
Since these 6 pairs constitute ONLY 1 LINE, we must subtract 5 of these 6 pairs from our total of 45:
45-5 = 40.
The correct answer is C.
Last edited by GMATGuruNY on Thu Jun 11, 2015 3:00 am, edited 1 time in total.
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Hi gmat_guy666,
Another way to think about this set-up is to 'map out' the options (without physically drawing them all).
I'm going to call the first 6 points A, B, C, D, E and F and the 4 collinear points G,H,I,J
Point A can form a line with any of the other 9 points = 9 lines
Point B can also form a line with any point (but already formed a line with point A, so we can't count that line twice) = 8 lines
Point C already formed lines with Point A and Point B.....= 7 lines
Point D = 6 lines
Point E = 5 lines
Point F = 4 lines
+ 1 more line formed by GHIJ....
9+8+7+6+5+4+1 = 40 lines
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
Another way to think about this set-up is to 'map out' the options (without physically drawing them all).
I'm going to call the first 6 points A, B, C, D, E and F and the 4 collinear points G,H,I,J
Point A can form a line with any of the other 9 points = 9 lines
Point B can also form a line with any point (but already formed a line with point A, so we can't count that line twice) = 8 lines
Point C already formed lines with Point A and Point B.....= 7 lines
Point D = 6 lines
Point E = 5 lines
Point F = 4 lines
+ 1 more line formed by GHIJ....
9+8+7+6+5+4+1 = 40 lines
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
