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If alice goes first

Alice can be 3 choices

Bob can have 2 choices

Carol can have 1 choice

In other words, it's just 3!

Alice, Bob, Carol:

3 ways to choose the person for the first job, then two ways, and finally one way.

3 * 2 * 1 = 6 ways.

Hint: For every quadrilateral there are 4 right triangles that can be created

People are unique! Don't forget that CPhill .

OOPSIES! I read the question wrong. It said how many right triangles. I think it is just manual labor from here on out.

Ah....thanks, tertre....I forgot  that  we  need to consider  the people as well as the  offices.....thanks for the correct answer  !!!!

Very nice, AELN  !!!!!

Is it P(5,3)=60 ways.

We  have  5 offices  and  we  want  to  choose any 3 of them....so

C(5,3)  =  10  ways  to do this

This is the  same  as  taking any set  of  three elements  from 5 =  10 sets

Here is a cheaty way to do it:

You can count the total number of ways 3 points get selected out of the 6

\({6 \choose 3}=20\) (Order does not matter)

Now, there are some cases that cannot be counted (straight lines)

DQC and APB are straight therefore are not triangles.

So 20 - 2 = 18? Counting and probability are my weakness I'm not sure if this is right.

A  =  30°

D                2

1

C          1                 B  =  60°

BC  =  1

AC  =   √3

If   BD  = √2

Then  DC  =   √[ (2)  - 1 ] =  1

The  area of  triangle  ABC  = (1/2) BC)(AC)  =  (1/2)(1)(√3) =  √3/2

Triangle  DCB  will  have  a max  area  when BD  = √2

And  its area  =  (1/2)(CD)(CB)  = (1/2)(1)(1)  = 1/2

So......the probability  that  BD < √2   will occur  when P  falls inside triangle  DCB

And this probability is     (1/2)  / (√3) /2  =   1 / √3

These numbers represent your cards if an ace = 1

1   1   1   1   2   2   2   2   3   3   3   3

____________________________________________________________

Plan:

Probability = Wanted Cases divided by Total Cases

Solve:

Total cases = 

\({12 \choose 4}\) because it does not matter how you distribute the cards.

12!/(4!*8!) = 495

Wanted cases (order does not matter):
Only once case

So  \(\frac{1}{495}\)? I'm not so sure about this one...

Umm... I got the (a+b)^2, its (x+3)^2, how would I continue Impasta. Or someone else?

If you really want to be ahead of the curve, take advantage of AoPS classes. I'm recommending them for their level of depth.

I'm glad you want to learn 

Yes please don't just give an answer to me, If you are, at least give an explanation too!