AngrySesame

There are \(10^4=10,000\) ways to choose M, A, T, and H. There is only one way to choose M, A, and H and two ways to choose T in MATHCOUNTS. 

That means that there are \(1 \cdot 1 \cdot 1 \cdot 2 = 2\) ways to spell MATH. Then, the probability of getting MATH is

\(\dfrac{2}{10,000} = \boxed{\dfrac{1}{5,000}}.\)

And yes, HeWhoShallNotBeNamed (@Voldemort)s' answer was correct.

So, I got something for the first section!

There are three possible categories of arrangements for \((a, b, c)\). They are:

• \((a, b, c)\) are equal.
• \((a, b, c)\) are all different.
• Two variables are equal, one is not.

There are \(6\binom{n}{3}\) arrangements when all the variables are different. \(\binom{n}{3}\) accounts for all the arrangements regardless of order, but we must multiply by 6 to get our final answer.

There are \(n\) arrangements when all the variables are the same. 

There are \(3n(n - 1)\) ways to compute the arrangements when two variables are the same, and one is not equal.

Now, I need to figure out b; which I cannot.

Gladly!