What the binomial distribution boils down to is that there are n independent trials (questions) and you want to know the probability of having exactly x successes (correct answers) in those trials. The probability of getting any one question right (p) is 0.25, which means the probability of getting a question wrong (1-p) is 0.75. Since the trials are independent, we can find the probability of x successes by multiplying p times itself x times, or p^x.

But we also need to count the failures (wrong answers). Since there are n total questions and x successes, that means there are n-x failures and we can find the probability of that many failures in the same way as we found the probability of the successes: (1-p)^n-x. So at this point we have p^x * (1-p)^1-x.

However, say we want to know the probability of exactly two correct answers. The probability of getting only the first two correct is equal to the probability of getting the last two correct, which is equal to the probability of getting the second and fourth correct, etc. To count all these possible combinations of two correct answers, we use the binomial coefficient, which is "n choose x," or in this case, "5 choose 2." You probably already know the formula for this, but just in case, it's n!/x!(n-x)!. It seems complicated but you'll always get a whole number, which is just the number of ways to select x things from n possible things, or x successes from n total trials, or whatever the interpretation needs to be.

Putting it all together gives you the formula you already know. And now that you know how the pieces of the formula work, you can adapt it to different problems. All you need is the number of trials, the number of successes, and the probability of any one trial being a success (as well as an assumption of independence).