rakeshd347 wrote:If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, How many triangles can be formed whose vertices are among the 9 points?
A. 20
B. 30
C. 40
D. 70
E. 90
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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 free 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 = [spoiler]70 = D[/spoiler]
Cheers,
Brent
