PS Question # 1 -How many different triangles can be formed?

[sumit.sinha]

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - C

[shovan85]

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - Soon

Total number of available points = 6
Total number of points needed to form a triangle = 3

Thus, total number of triangles can be formed = C(6,3) = 6!/(3!*3!) = 4*5 = 20

As there in no 3 points co-linear we dont have to subtract anything.

IMO C

[sumit.sinha]

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - Soon

Total number of available points = 6
Total number of points needed to form a triangle = 3

Thus, total number of triangles can be formed = C(6,3) = 6!/(3!*3!) = 4*5 = 20

As there in no 3 points co-linear we dont have to subtract anything.

IMO C

Have you made sure that you are not counting the same triangle twice?

[shovan85]

I don't know. According to formula it worked.

By considering AB as base AFB the red triangle, ADB the green one, ABC the yellow one, and ACF the blue-yellow one, I got 4 triangles.

Please help me see if I am recounting any triangle more than once.

[sumit.sinha]

I don't know. According to formula it worked.

By considering AB as base AFB the red triangle, ADB the green one, ABC the yellow one, and ACF the blue-yellow one, I got 4 triangles.

Please help me see if I am recounting any triangle more than once.

With base AB: you missed the triangle ABE and triangle ACF does not have base AB.

But more importantly my doubt has cleared, I was not counting the triangles like ACF.
With base as AB - 4 different triangles
With base as BC - 3 different triangles
With base as CD - 3 different triangles
With base as DE - 3 different triangles
With base as AE - 2 different triangles
the five i was missing were:
ACF, ADF,BEF,BDF and CEF

Ok. i guess i am going to use the formula next time around. Much faster. Thanks.

[anshumishra]

I don't know. According to formula it worked.

By considering AB as base AFB the red triangle, ADB the green one, ABC the yellow one, and ACF the blue-yellow one, I got 4 triangles.

Please help me see if I am recounting any triangle more than once.

I don't think there should be any repetitions.
sumit probably is thinking whether there are multiple ways in which 3 points are counted twice (for e.g, [A,B,C] and [B,A,C]).
However, the combination formula takes care of that, so when we select r items out of n, all the nCr ways are unique sets.

If it would have been nPr, then the counting will make difference as the arrangements [A,B,C] and [B,A,C] would have been considered differently.

In our case [A,B,C] and [B,A,C] are same, they make same triangle. So, Shovan's approach looks good to me (unless I am missing something here).

Thanks

[shovan85]

With base AB: you missed the triangle ABE and triangle ACF does not have base AB.

No man I did not miss ACF triangle. I have highlighted it in blue in the picture.

And the reason I did not consider ABE as I can see it is similar to ABC and plus it is using two sides of the polygon. Thus is bound to get repetitive.

Hope OA is 20

[sumit.sinha]

With base AB: you missed the triangle ABE and triangle ACF does not have base AB.

No man I did not miss ACF triangle. I have highlighted it in blue in the picture.

And the reason I did not consider ABE as I can see it is similar to ABC and plus it is using two sides of the polygon. Thus is bound to get repetitive.

Hope OA is 20

yes, it is.

[diebeatsthegmat]

Answer Choices:
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Source - Princeton CAT5

OA - Soon

Total number of available points = 6
Total number of points needed to form a triangle = 3

Thus, total number of triangles can be formed = C(6,3) = 6!/(3!*3!) = 4*5 = 20

As there in no 3 points co-linear we dont have to subtract anything.

IMO C[/quote
very nice, i love this solution. thanks!