AnExtremelyLongName

That is true! Technically, if there are 0 blocks, the blocks can be evenly distributed 

Not sure if this helps, but graphing may work. Try setting up circle inequalities?

This is quite the same as the Alice problem.

First dial can have all ten cases

Second dial has to be different than the first so it has to have 9 cases

Third dial has to be different than the first and second so it has to have 8 cases.

10 * 9 * 8 = 

My parents won't allow me to take AoPS classes because they cost like hundreds of dollars. I have to rely learning from online resources :/

Calculator User is in the same situation lol.

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!

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.

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.