Problem Solving

BTGmoderatorLU wrote:Source: e-GMAT

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

The OA is B.

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, we have a video on calculating combinations (like 5C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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)
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So, the total number of triangles = 40 + 30
= 70

Answer: B

Cheers,
Brent

BTGmoderatorLU wrote:Source: e-GMAT

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

The OA is B.

We can let line A be the line with 5 points and line B the line with 4 points. So we can have a triangle that is formed from choosing 1 point on line A and and 2 points on line B, or, 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, there are 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 all 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 all 3 points from the line with 4 points is 4C3 = 4. Therefore, there are a total of 84 - 10 - 4 = 70 ways forming a triangle.

Answer: B