I've just recently learned about permutations and combinations, but a thought came up to me and it might just be my brain not being particularly as bright as it usually is, but how would I go about finding out the number of different ways that one can fill out a test (in terms of answers)? For example, let's say that there's a 20 question test, and a person can answer each question with either A, B, C, or D. How many combinations of answers are there? Would I do $\frac{20!}{(20-5)!}$ (permutation), $\frac{20!}{5!(20-5)!}$ (combination, which I don't think is correct, but you never know), or something like $4^{20}$?
Help would be very much appreciated. Thanks!
The way that I usually deal with permutations and combinations is by thinking about where the formulas come from in the first place.
For example, if you're interested in how many different pairs you can have out of the triple $A, B, C$, you're looking for all the sets of the form $\{x, y\}$. There are $3$ choices for $x$, and then $2$ choices for $y$. This gives us $3 \times 2 = 6$ potential pairings. But, we are dealing with sets, and we notice that $\{x, y\} = \{y, x\}$, so the counting about needs to be divided by two to account for the ordering of $x$ and $y$ not mattering. Hence the number of pairs is $6 / 2 = 3$, which you can verify by drawing them out. This corresponds to the formula $$ \frac{3!}{2!(3-1)!} = \frac{3!}{2!} = 3 $$ which you know to be the "choose function".
This is the same if we were choosing sets of size three out of the four elements $A, B, C, D$. We are after sets of the form $\{x, y, z\}$, of which there are $4 \times 3 \times 2$ choices initially. But how many of them are the same? Well, the following sets are all equal: $$ \{x, y, z\}, \{x, z, y\}, \{y, x, z\}, \{y, z, x\}, \{z, x, y\}, \{z, y, x\}. $$ There is six sets that are all the same, which we note is the same as $3!$ (this is not a coincidence, convince yourself that this will always be the case with these set-like questions). Thus, the number of distinct sets that one can choose of size three out of four elements is $$ \frac{4\times 3\times 2}{3!} = \frac{4!}{3!\cdot1!} = \frac{4!}{3!(4-3)!} = 4. $$ Notice that again this is precisely the form of the choose function.
Now, what about your question? Well, assuming that an answer must be chosen, there are four choices for each question, and twenty questions. In the first question you have $4$ choices, in the second question you have $4$ choices, in the third question you have $4$ choices, ..., in the twentieth question you have $4$ choices, so you have $$ 4 \times 4 \times 4 \times \cdots \times 4 = 4^{20} $$ choices. In this case, there are no equivalent choices, so you don't need to divide by anything and you're done.