D01-40

it is a regular polygon with 7 sides. Regular polygons have equal sides, and equal angles.

for the sake of understanding, draw a 7 sides polygon and start drawing triangles and quadrilaterals.
we clearly see that from one angle, we can draw lines and form 4 triangles. How many triangles can be? well, we have 7 sides, and from 7 sides, we can have 4 triangles
4C7 = 7!/4!3! = 5*6*7/1*2*3 = 35
now, draw quadrilaterals. From each angle, we can draw lines and form 3 quadrilaterals. How many quadrilaterals can be drawn? 7 sides, 3 quadrilaterals...
3C7 = which is equal to 4C7
now we have 35+35 = 70.

I think this is a high-quality question and I agree with explanation.

I drew the lines and ended up with 5 triangles (not 4) and 4 quadrilaterals (not 3): Red, Purple, Green, and Blue.
Check the images that I drew.
Please, point out, where is the mistake.

Thank you.

As an example I'll show you 2 triangles that you missed:You are missing triangles and quadrilaterals that can be formed from other vertices.

I think you missing 1 triangle and 1 quadrilateral, Please, see your quote:


In my images I stated that you can form 5 triangles from one angle, and 4 quadrilaterals from one angle. While You formed 4 triangles and 3 quadrilaterals.

So shouldn't be the solution:

7C5+7C4=21+35=56 different triangles and quadrilaterals can be formed

Sorry I cannot draw all 35 triangles and 35 quadrilaterals but the logic in my solution explains why it is so. Any 3 distinct points from 7 will form a triangle and any 4 distinct points from 7 will form a quadrilateral.

Bunuel, if you could explain this method just algebraically for square and pentagon (how many triangles can be formed), so I could better understand your logic.

Thanks

Below are cases for square. We can make only 1 (4C4 = 1) quadrilateral using the vertices of a square and we can make 4 triangles (4C3 = 4).

TRIANGLES
If you choose any 3 of the 7 vertices, you can connect them with lines to create a unique triangle.
So, the question becomes "In how many different ways can we select 3 vertices from 7 vertices?"
Since the order in which we select the 3 vertices does not matter, we can use COMBINATIONS.
We can select 3 vertices from 7 vertices in 7C3 ways.
7C3 = 35

ASIDE: we have a video on calculating combinations (like 7C3 and 7C4) in your head - see below


QUADRILATERALS
If you choose any 4 of the 7 vertices, you can connect them with lines to create a unique quadrilateral.
So, in how many different ways can we select 4 vertices from 7 vertices?
Since the order in which we select the 4 vertices does not matter, we can use COMBINATIONS.
We can select 4 vertices from 7 vertices in 7C4 ways.
7C4 = 35

So, the TOTAL number of triangles and quadrilaterals possible = 35 + 35 = 70

Answer: E

Cheers,
Brent

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I think this is a high-quality question and I agree with explanation. I don't think this is a 600 level question, maybe a medium 600-700 level one.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I think this is a high-quality question and I agree with explanation.

Don't we need to factor in for instances where the lines are co-linear? In a sense, they're on the same side of the polygon

The point is that no two sides in a seven-sided regular polygon are parallel.

I think this is a high-quality question. Is this question a part of the GMAT focus syllabus?