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# M03 #23

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Manager
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04 Oct 2008, 03:55
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?

(A) 168
(B) 196
(C) 316
(D) 364
(E) 455

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Would anyone know how to solve for this?
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04 Oct 2008, 04:27
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If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?

(C) 2008 GMAT Club -
168
196
316
364
455
Would anyone know how to solve for this?

A triangle needs 3 distinct/non-collinear points to form the three vertices.

Line 1 has 7 points and Line 2 has 8. We cannot select all 3 points from the same line to form a triangle.

we have 2 options. Select 2 points from Line1 and 1 from Line2 (OR) Select 1 point from Line1 and 2 from Line2.

in other words, 7C2*8C1 + 7C1*8C2 (we multiply when we have "AND" and add when we have "OR")

= 364
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14 Jul 2010, 06:34
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there are 7+8=15 points in all.
"combination" of any 3 points taken at a time will give us a triangle so 15c3=455
BUT if all the 3 points we choose lie on the same line then they will be co-linear and hence wont form a triangle.
SO we subtract all such combination's i.e (7c3 + 8c3)= 91
therefore 455-91=364
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14 Jul 2010, 08:45
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Sometimes it's not bad to forget about combinations and permutations and respective formulae. Each point out of 7 on the 1st line can be joined with 8*7 pairs on the 2nd, and any out of 8 the 2nd we join with 7*6 pairs on the 1st.
Just don't forget to divide the sum by 2, otherwise each triangle will counted twice. (7*8*7+8*7*6)/2=364.

Last edited by nvgroshar on 19 Jul 2011, 07:54, edited 1 time in total.
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15 Jul 2010, 06:54
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Check your calculation on Step 2: Result is 28. This is multiplied by 7 to get 196.
So, total comes out to be 168 + 196 = 364
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24 Jan 2012, 19:05
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vitaliti23 wrote:
I have a quick question on this problem. When I saw, I immediately did this. Assuming we start on the line with 8 points, I took the number 8. I then multiplied that by 7 (the options on the second line). I then multiplied that number by 7 (the number of options left on the first line). I came out with 392. What did I do wrong? How was my thinking incorrect on this?

Hello, and welcome to GMAT Club.

1. When you pick 2 points from the line with 8 points by 8*7 you'll count twice different pairs of two points from this line. Correct way of choosing 2 different points from 8 would be $$C^2_8=28$$ which is half of 8*7=56, so to get rid of these duplication you should divide 56 by 2!.

Consider this: how many different pairs of two letters are possible from ABC - (AB), (BC) and (AC), only 3 ($$C^2_3=3$$). If we do your way we'll get 3*2=6 twice as many.

2. Another flaw is that you are only considering the triangles which have two vertices on the line with 8 points and 1 vertex on the line with 7 points, but opposite case is also valid: two vertices on the line with 7 points and 1 vertex on the line with 8 points.

Below are two main approaches for this problem.
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(A) 168
(B) 196
(C) 316
(D) 364
(E) 455

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 8 points and the third vertex on the line with 7 points --> $$C^2_8*C^1_7=28*7=196$$;
With two vertices on the line with 7 points and the third vertex on the line with 8 points --> $$C^2_7*C^1_8=21*8=168$$;

Total: $$196+168=364$$.

Approach #2:

All different 3 points out of total 8+7=15 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364$$ (where $$C^3_8$$ and $$C^3_7$$ are # of different 3 collinear points possible from the line with 8 points and the line with 7 points, respectively).

Hope it helps.
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14 Jul 2010, 06:20
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I agree with the solution . It is D.

When I looked at the problem. I thought I should use permutations rather than combinations. My solution was not one of the answer choices and hence switched to Combinations.

Could somebody explain , why combinations should have been the choice in lay man terms.

Thanks a lot
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14 Jul 2010, 06:49
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Permutation is used where arrangements matter. Combination is used here since the order of selection does not matter.
i.e. with 3 points chosen (1 point from one line and 2 points from the other parallel line), only a unique triangle can be formed using the 3 points.
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15 Jul 2010, 04:22
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amitdgr wrote:
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?

(C) 2008 GMAT Club -
168
196
316
364
455
Would anyone know how to solve for this?

A triangle needs 3 distinct/non-collinear points to form the three vertices.

Line 1 has 7 points and Line 2 has 8. We cannot select all 3 points from the same line to form a triangle.

we have 2 options. Select 2 points from Line1 and 1 from Line2 (OR) Select 1 point from Line1 and 2 from Line2.

in other words, 7C2*8C1 + 7C1*8C2 (we multiply when we have "AND" and add when we have "OR")

= 364

VP - Would you mind walking me through the math on this. I had the same setup, but got a different answer.

Step 1: Number of combinations of selecting 2 points on Line 1, and one point on Line 2: 7C2 = (7!) / (2!)(7 - 2)! = (7!) / (2!)(5!) = (7*6) / (2) = 21. Now multiply that by 8C1 = 8 to get 168.
Step 2: Number of combinations of selecting 2 points on Line 2, and one point on Line 1: 8C2 = (8!) / (2!)(8 - 2)! = (8!) / (2!)(6!) = (8*7) / (2) = 91. Now multiply that by 7C1 = 7 to get 91.
Step 3: 168 + 91 = 259

What am I missing here? Thanks in advance.
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18 Jul 2011, 08:58
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If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?

(A) 168
(B) 196
(C) 316
(D) 364
(E) 455

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Would anyone know how to solve for this?

$$7 * C^8_2+8 * C^7_2$$

OR

$$C^{(8+7)}_3-C^7_3-C^8_3$$
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19 Jul 2011, 18:47
I actually drew 2 parallel lines and marked 7 points on the first line (L1) and 8 on the second line (L2).

Take the first point on L1. You can connect it to 8 different points on L2.That means for P1 in L1 you get 7+6+5+4+3+2+1 = 28 different triangles.

There are 7 points so in total you get 7*28 = 196 different triangles.

Repeat the same process with L2. You get 8*21 (6+5+4+3+2+1) = 168 different triangles.

Add the two up and you get 364 different triangles. Hence answer is D.

This is a somewhat long method and took me about 3 minutes to do.
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20 Jul 2011, 21:50
why cant we choose one point from each line and then choose one from the remaining 13

ie 7C1*8C1*13C1 and divide by 2

But i cant understand y the triangles are repeating
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24 Jan 2012, 18:21
I have a quick question on this problem. When I saw, I immediately did this. Assuming we start on the line with 8 points, I took the number 8. I then multiplied that by 7 (the options on the second line). I then multiplied that number by 7 (the number of options left on the first line). I came out with 392. What did I do wrong? How was my thinking incorrect on this?
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24 Jan 2012, 21:32
Bunuel wrote:
vitaliti23 wrote:
I have a quick question on this problem. When I saw, I immediately did this. Assuming we start on the line with 8 points, I took the number 8. I then multiplied that by 7 (the options on the second line). I then multiplied that number by 7 (the number of options left on the first line). I came out with 392. What did I do wrong? How was my thinking incorrect on this?

Hello, and welcome to GMAT Club.

1. When you pick 2 points from the line with 8 points by 8*7 you'll count twice different pairs of two points from this line. Correct way of choosing 2 different points from 8 would be $$C^2_8=28$$ which is half of 8*7=56, so to get rid of these duplication you should divide 56 by 2!.

Consider this: how many different pairs of two letters are possible from ABC - (AB), (BC) and (AC), only 3 ($$C^2_3=3$$). If we do your way we'll get 3*2=6 twice as many.

2. Another flaw is that you are only considering the triangles which have two vertices on the line with 8 points and 1 vertex on the line with 7 points, but opposite case is also valid: two vertices on the line with 7 points and 1 vertex on the line with 8 points.

Below are two main approaches for this problem.
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?
(A) 168
(B) 196
(C) 316
(D) 364
(E) 455

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 8 points and the third vertex on the line with 7 points --> $$C^2_8*C^1_7=28*7=196$$;
With two vertices on the line with 7 points and the third vertex on the line with 8 points --> $$C^2_7*C^1_8=21*8=168$$;

Total: $$196+168=364$$.

Approach #2:

All different 3 points out of total 8+7=15 points will create a triangle EXCEPT those 3 points which are collinear.
$$C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364$$ (where $$C^3_8$$ and $$C^3_7$$ are # of different 3 collinear points possible from the line with 8 points and the line with 7 points, respectively).

Hope it helps.

Thanks so much for your help!
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20 Jul 2012, 05:27
If one of two parallel lines has 7 points on it and the other has 8, how many triangles can be drawn using the points?

(A) 168
(B) 196
(C) 316
(D) 364
(E) 455

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Would anyone know how to solve for this?

7c1*8c2+8c1*7c2

D
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20 Jul 2012, 07:26
sjgudapa wrote:
I agree with the solution . It is D.

When I looked at the problem. I thought I should use permutations rather than combinations. My solution was not one of the answer choices and hence switched to Combinations.

Could somebody explain , why combinations should have been the choice in lay man terms.

Thanks a lot

I would suggest to forget about combinations, permutations, factorials...
Think of the real process and translate the steps into simple arithmetic operations:

For a triangle, we need three different vertices, which are not all three on the same line.
Choose one vertex, say A, from the points on the line with the 7 points and the two other vertices, say B and C, from the points on the line with the 8 points.
For A we have 7 choices, and for each of these, for B 8 choices, then for C 7 choices. This will translate into 7 * 8 * 7 choices, but we must divide by 2, because choosing first B, then C, will produce the same triangle as choosing first C and then B. So, we can get 7 * 8 * 7/2 distinct triangles.

Repeat the above process for triangles with vertex A on the line with 8 points, and vertices B and C on the line with 7 points.
This will give you 8 * 7 * 6 / 2 triangles.

All the triangles chosen above are distinct. Therefore, the total number of triangles is 7 * 4 * 7 + 4 * 7 * 6 = 28(7 + 6) = 28 * 13 <--- this should end in 4, and luckily, we have just one answer which ends in 4. If there would have been more than one answer which ends in 4, we should have carried out the multiplication or do some other estimate.
So, the correct answer is 364.

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20 Jul 2012, 11:21
Followed the approach of choosing the points as has been mentioned in the previous posts.

1. 2 points on line1 having 7 points and one point on the line2 having 8 points = 7C2*8C1

2. 2 points on line2 having 8 points and one point on the line1 having 7 points = 8C2*7C1

Total = 7C2*8C1 + 8C2*7C1
= 21*8 + 28*7
= 7 (3*8 + 4*7)
= 7 * 52
= 364

Hence, D.
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26 Jul 2012, 16:52
To create a triangle we need to select 3 points, 2 from 1st line & one from the 2nd line.

8C2*7C1(we are selecting 2 points from line having 8 points & 1 point from line having 7 points)
+
7C2*8C1(we are selecting 2 points from line having 7 points & 1 point from line having 8 points)

= 196 + 168 = 364
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23 Jul 2013, 03:36
thanks for the solution...
Re: M03 #23   [#permalink] 23 Jul 2013, 03:36
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# M03 #23

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