GMAT Math

PAB2706 wrote:There are eight points marked on the circle. how many different pentagons and different triangles can be drawn using these eight points. For each pentagon that is drawn, a blue disc is put in a bag and for each triangle that is drawn a red disc is put in the bag. what the the probability of choosing a blue disc from the bag after all the possible pentagon and triangle are drawn.

Are are two approaches:

The first (longer) approach: apply counting rules
Premise: any selection of 5 points will define a unique pentagon.
So, we can reword the question as: In how many ways can we select 5 points?
Well, since the order of the selected points does not matter, this is a combination question.
So, the number of ways to select 5 points from 8 points is 8C5 ("8 choose 5")

Similarly, any selection of 3 points will define a unique triangle.
So, we can reword the question as: In how many ways can we select 3 points?
Once again the order of the selected points does not matter, this is a combination question.
So, the number of ways to select 3 points from 8 points is 8C3 ("8 choose 5")

Now we could calculate 8C5 and 8C3, but there's a useful rule that says nCk = nC(n-k)
For example, 11C2 = 11C9 and 10C6 = 10C4
This means that 8C5 = 8C3, so the number of possible pentagons is equal to the number of possible triangles.
In other words, the number of blue discs in the bag is equal to the number of red discs.
So, P(select a red disc) = 1/2

PAB2706 wrote:There are eight points marked on the circle. how many different pentagons and different triangles can be drawn using these eight points. For each pentagon that is drawn, a blue disc is put in a bag and for each triangle that is drawn a red disc is put in the bag. what the the probability of choosing a blue disc from the bag after all the possible pentagon and triangle are drawn.

The second (faster) approach: apply logic
Recognize that if we select 5 points, then that defines a unique pentagon, and the remaining 3 points define a unique triangle.
So, we can see that for every unique pentagon there is one unique triangle.
So, P(select a red disc) = 1/2