A professor writes 40 discrete mathematics multiple choice questions each with three possible answers, a), b), or c). If there are absolutely no constraints on how many questions can have answer a) or answer b) or answer c) and the questions can be positioned in any order, how many dierent answer keys are possible?
Would I solve this by just doing 40$^3$
No, $3^{40}$ It seems a fancy way of saying "you have 40 choices, you may choose from 3 outcomes for each"
The axiom of choice: Given a decision with $m$ outcomes and another that will always have $n$ outcomes regardless of what you do for the first, there are $mn$ possible outcomes.
You have a choice, 3 for question 1 and 3 for question 2, this gives $3^2$ for the first 2, you then have 3 for the third, giving $3^3$.... and the 40th gives $3^{40}$.
If I can choose 1 of 10 things, and I do it 4 times there are 10 x 10 x 10 x 10 ways, or $10^4$, which is 10,000, if those were numbers the number of possible 4 digit numbers is 10,000, 0000 to 9999 inclusive.
If I have a set of n things, and I choose them in a certain order, there are n for the first choice $n-1$ for the second, $n-2$ for the third.... which is where factorial comes from.