Question: In a triangle ABC, there are 5, 6, and 4 points, different from the vertices A, B, and C of the triangle, on the three sides AB, BC, and CA, respectively. How many distinct triangles can be drawn using the 18 points (including the vertices A, B, and C)?

1. 111
2. 705
3. 816
4. 910
5. 955

Answer:
Solution with Explanation:
Approach Solution (1):

First, let’s pick 3 points out of 18
We get \(\frac{18*17*16}{3*2*1}=816\)

But, any three points exclusively from AB, or BC, or CA cannot form a triangle

There are a total of 5 + 2 = 7 points on AB. Pick three from seven, and we get \(\frac{7*6*5}{3*2*1}=35\)
There are total of 6 + 2 = 8 points on BC. Pick three from eight, and we get \(\frac{8*7*6}{3*2*1}=56\)
There are total of 4 + 2 = 6 points on CA. Pick three from six, and we get \(\frac{6*5*4}{3*2*1}=20\)

Final answer: 816 – 35 – 56 – 20 = 705

Correct Option: B

Approach Solution (2):

First, you could have the case in which you pick one point from each of the 3 sides of triangle
Second, you could pick 2 points from one side and a 3rd from the remaining two sides

However, you would have to account for over counting as well, since the corner vertices will be counted as part of two sides
On the other hand, it is much easier to just subtract out the unfavorable combinations

Since we need 3 vertices/points to make a triangle, first we can find every unique way to make a combination of 3 points out of 18 available points:

(Total number of unique ways to make combinations of 3 points out of a total of 18 available points)
(Unfavorable combinations in which the 3 points are collinear and can’t represent a triangle)

This is equal to:
18C3

Number of ways to have a group of 3 linear points from side AB: 7C3
Number of ways to have a group of 3 linear points from side BC: 8C3
Number of ways to have a group of 3 linear points from side AC: 6C3

When you calculate the factorials, you get:
816 – (35 + 56 + 20) = 705

Correct Option: B

Approach Solution (3):

There are a total of 18 points from which we need three non-collinear ones
\(\frac{18*17*16}{3*2*1}=3*17*16=51*16=800+16=816\)

Now we need to eliminate some C, D, and E are out
We are only eliminating the ones with three points all from the same side. That’s not the majority. A is out

Correct Option: B

“In a triangle ABC, there are 5, 6, and 4 points, different from the vertices A, B, and C of the triangle, on the three sides AB, BC, and CA, respectively. How many distinct triangles can be drawn using the 18 points (including the vertices A, B, and C)?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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