# Probability Question (1 Viewer)

#### AreYouAlright?

##### Actuarial Co-op 2006

Twelve points are arranged in order around a circle
a) How many triangles can be drawn with these points as verticies? ( I can do this )
b) How many pairs of such triangles can be drawn if the vertices of the two triangles are distinct? (I can do this too )
c) In how many such pairs will the triangles i) Not overlap ii) Overlap????

#### 香港!

##### Member
aye I hate probability!!

#### Mill

##### Member
I'm a little busy atm but I found this question too interesting to pass up. However I only took a minute to think about it so I am definitely not sure that what I am about to type is correct.

(ii) ANSWER TO PART 2 DENOTED AS X (you will see why in a minute).

(iii) For you to have two triangles with distinct vertices, you must have 6 distinct vertices, so let's consider them.

Assign one vertex arbitrarily to the first triangle.

Now, in order for the triangles to NOT overlap, the 2 remaining vertices MUST be the 2 points next to our arbitrarily chosen original point.

You have a 1/10 chance of selecting those particular 2 points (there is 5C2 = 10 totals ways to choose 2 points from the 5).

Therefore if you have two triangles made from 6 distinct vertices, 10% of them will not overlap. So your answer would be 1/10 . X

#### AreYouAlright?

##### Actuarial Co-op 2006
Thanks for the attempt Mill, however that is not the correct answer. I have no idea how you get the answer to this question.

The numerical answers for this question are as follows:

a) 220
b) 9240
c)i. 2772
ii. 6468

Last edited:

#### Mill

##### Member
Ok I have worked out where my mistake was.

The basic idea of how I was working is fine.

You arbitrarily select an original point, and there are then 5C2 = 10 ways to select 2 remaining points.

However, THREE of these 10 ways will give you a triangle that does not overlap (ie. the points are successive), not one as I was originally thinking.

Draw a diagram to see this clearer.

If you have 6 points in a circle, and choose your first one, you can then choose two to the left OR two to the right OR one on either side, ie. there are THREE ways!!

Hence, the number of non-overlapping triangles is 3/10 . X.

This agrees with the answer posted by the OP.

Sorry I took so long to get back to this and fix the error up. Hope that helps.

#### tywebb

##### dangerman
You can do it like this:

(a) $\bg_white ^{12}C_3=220$

(b) $\bg_white \textstyle\frac{^{12}C_3\times{}^9C_3}{2}=9240$

(c)(i) $\bg_white \textstyle\frac{12\times({}^9C_3\times1+{}^8C_3\times2+{}^7C_3\times3+{}^6C_3\times4+{}^5C_3\times5+{}^4C_3\times6+{}^3C_3\times7)}{2}=2772$

(c)(ii) $\bg_white 9240-2772=6468$

#### hscgirl

##### Active Member
You can do it like this:

(a) $\bg_white ^{12}C_3=220$

(b) $\bg_white \textstyle\frac{^{12}C_3\times{}^9C_3}{2}=9240$

(c)(i) $\bg_white \textstyle\frac{12\times({}^9C_3\times1+{}^8C_3\times2+{}^7C_3\times3+{}^6C_3\times4+{}^5C_3\times5+{}^4C_3\times6+{}^3C_3\times7)}{2}=2772$

(c)(ii) $\bg_white 9240-2772=6468$
wait, how can the answer (b) be larger than (a)??? because part b) is saying 'how many pairs of such triangles can be drawn if the vertices of the two triangles are distinct?', so it is referring to a subset of the triangles in part a) (which is only 220), right?
note: the maths behind the answer to both part a) and part b) makes sense to me, i just don't understand how mathematically b) is a larger number.

ik this q is from 2005 but someone posted on this on wednesday and i'm currently doing this q for my hw and i'm so confused why b) is larger than a), so someone plz help

#### hscgirl

##### Active Member
does anyone know the reason why??

#### synthesisFR

##### Well-Known Member
You can do it like this:

(a) $\bg_white ^{12}C_3=220$

(b) $\bg_white \textstyle\frac{^{12}C_3\times{}^9C_3}{2}=9240$

(c)(i) $\bg_white \textstyle\frac{12\times({}^9C_3\times1+{}^8C_3\times2+{}^7C_3\times3+{}^6C_3\times4+{}^5C_3\times5+{}^4C_3\times6+{}^3C_3\times7)}{2}=2772$

(c)(ii) $\bg_white 9240-2772=6468$
bro responded 18 years later