Probability using Combinations I am confused on how this works. Normally, probability is:
$$P = \frac{\text{Number of successes}}{\text{Number of total trials}}$$
For a problem like:

If you flip a fair coin $8$ times, what is the probability of getting exactly $3$ out of $8$ heads? 

The solution first depicts how many ways there are to choose $3$ heads. With:
$$\binom{8}{3}$$
Then what will the denominator be? Moreover what is this method?
 A: Numerator: how many ways can you get exactly 3 heads and 5 tails? This is a matter of choosing which three of the eight trials get heads: $\binom{8}{3}$.
Denominator: how many possible outcomes are there? For each trial there are two outcomes, so there are $2^8$ possible outcomes for eight flips.

Addendum: As noted in the other answer, the only reason we are allowed to use combinatorics in this manner for this question is because every possible outcome of 8 flips has the same probability. From this observation, the problem naturally becomes a combinatorics problem because we are counting outcomes in the manner you outlined in your question.
A: After you flip the coin $8$ times you will have an 8-letter word in front of you consisting of "H"'s (for Heads) and "T"'s (for Tails). Now ask yourself:


*

*How many such words are there? You are only interested only in the number of "H"'s in the words and not in their exact place, so there are $$\dbinom{8}{0}+\dbinom{8}{1}+\ldots+\dbinom{8}{8}=\sum_{k=0}^{8}\dbinom{8}{k}=(1+1)^8=2^8$$ where the last equation is due to the binomial identity (or theorem). This number gives you the denominator (number of different possible outcomes).

*How many of these words have exactly $3$ "H"'s? You already have that this is equal to $\dbinom{8}{3}$.


Thus the required probability is equal - as you correctly know - to $$\frac{\text{Number of successes}}{\text{Number of total trials}}=\frac{\dbinom{8}{3}}{2^8}$$

Note however that this works because each possible result has the same probability. So I will not agree with your first sentence that "normally...". 
Now, combinatorics are used in order to count things in a more efficient way. In the hypothetical case that you have a lot (really lot) of spare time and a lot of fingers (>>20) you could count things using time and fingers and combinatorics would be useless. If that is not the case then you will have to use combinatorics in probability rather frequently.
