In a plane, there are two parallel lines. One line has 5 points and an

Question

In a plane, there are two parallel lines. One line has 5 points and another line has 4 different points. How many different triangles can we form from these 9 points?

A. 62
B. 70
C. 73
D. 86
E. 122

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EgmatQuantExpert · Accepted Answer

Solution

Given  :

•There are two parallel lines in a plane. 
 • One line has 5 points.
 •Another line has 4 points.

To find  : 
 • The number of different triangles using the 9 points on two lines.

Approach and Working out  :

A triangle is formed by 3 different points. 
Hence, we need to select 3 different points from 9 different points.
But, there is a catch.

• Can you form a triangle if all the three points lie on one line?? 
 • No.

Let us call the line having 5 points as L1 and line having 4 different points as L2.

Since three different points give only one triangle, this is a combination question.

Thus, the possible cases to form a triangle are: 
 • Select two points on L1 and one point on L2 
• Select two points on L2 and one point on L1.

Total ways to form the triangle=  ^5C_2 *  ^4C_1  +  ^4C_2 *  ^5C_1 
 • = 10*4+6*5= 40+30=70

Hence, option B is the correct answer.

Answer: B

styxnexus · Answer

9c3−4c3-5c3 = 70. Hence  option B.

aserghe1 · Answer

There are 2 parallel lines. 5 points on one line, 4 points on the other line. To make a triangle, you need 2 points from one line and 1 point from the other line.

# of triangles = (# of ways to select 2 out of 5 points) and (# of ways to select 1 out of 4 points) or (# of ways to select 1 out of 5 points) and (# of ways to select 2 out of 4 points)

# of ways to select 2 out of 5 points: 5C2=10
# of ways to select 1 out of 4 points: 4C1=4
# of ways to select 1 out of 5 points: 5C1=5
# of ways to select 2 out of 4 points: 4C2=6

# of triangles = 10*4+5*6 = 70

Answer: B

EgmatQuantExpert · Answer

Hey Everyone,

Official solution to the question has been posted.

Regards,
Ashutosh

zishu912 · Answer

There are two parallel lines.

One line has five points and another has 4 points in it.

So while selecting points we can't select all the three points from the same line, since it will be colinear
Hence we will select two points from one line and other from second line, which gives us:
5C2*4C1 + 4C2*5C1 = 70

BrentGMATPrepNow · Answer

There are two ways in which we can create a triangle. 
#1) Select 2 points from the 5-point line and select 1 point from the 4-point line.
#2) Select 2 points from the 4-point line and select 1 point from the 5-point line.

#1) Select 2 points from the 5-point line and select 1 point from the 4-point line. 
Take this task and break it into stages.

Stage 1 : Select 2 points from the 5-point line 
Since the order of the 2 selected points does not matter, we can use combinations. 
We can select 2 points from 5 points in 5C2 =  10  ways.

If anyone is interested, a video on calculating combinations (like 5C2) in your head can be found at the bottom of this post

Stage 2 : Select 1 point from the 4-point line. 
We can complete this stage in  4  ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in  (10)(4)  ways (=  40  ways)

#2) Select 2 points from the 4-point line and select 1 point from the 5-point line. 
Take this task and break it into stages.

Stage 1 : Select 2 points from the 4-point line 
We can select 2 points from 4 points in 4C2 =  6  ways.

Stage 2 : Select 1 point from the 5-point line. 
We can complete this stage in  5  ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in  (6)(5)  ways (=  30  ways)
-------------------------------------------------------------

So, the  TOTAL  number of triangles =  40  +  30  
= 70

Answer: B

Cheers,
Brent

RELATED VIDEO
 https://www.youtube.com/watch?v=05a0A7vjG8I

ScottTargetTestPrep · Answer

Solution:
 
We can let line A be the line with 5 points and line B the line with 4 points. So we can form a triangle by choosing 1 point on line A and 2 points on line B, or by choosing 2 points on line A and 1 point on line B. In the former case, we have 5C1 x 4C2 = 5 x 6 = 30 ways of forming such a triangle, and in the latter case, we have 5C2 x 4C1 = 10 x 4 = 40 ways of forming such a triangle. Therefore, we have a total of 30 + 40 = 70 ways to form a triangle.

Alternate solution:

The number of ways to choose 3 points from 9 is 9C3 = 9! / (3!6!) = (9 x 8 x 7) / (3 x 2) = 3 x 4 x 7 = 84. However, we can’t have all 3 points on the same line (since the 3 points can’t be collinear). The number of ways of choosing 3 points from the line with 5 points is 5C3 = 5! / (3!2!) = (5 x 4 x 3) / (3 x 2) = 5 x 2 = 10 and similarly, the number of ways of choosing all 3 points from the line with 4 points is 4C3 = 4. Therefore, we have a total of 84 - 10 - 4 = 70 ways forming a triangle.

Answer: B

mauja007 · Answer

total ways = no.of ways selecting 2 points from 5 and one point from 4 + no.of ways selecting 2 points from 4 and one from 5 
= 5c2 * 4 + 4c2 * 5
= 10 * 4 + 6*5
40 + 30 =70

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