AmyMath03

The answer is 316,251.

We can start with casework to get an idea:
Case one: Everyone (all fifty) vote: this is 53 choose 3. This is because we are employing the stars and bars method where the "stars" are the votes total, and the bars divide them up into 4 chunks for 4 candidates.

Case two: One person decides not to vote: this is 52 choose 3.

We can continue like this, and when we add the cases up we will end up with something like this:

${53\choose3}+{52\choose3}+{51\choose3}+\cdots+{4\choose3}.$

(that last case is when only one person votes, and we have 1 star and 3 bars and need to find 3 places to put 3 bars).

This looks familiar!

If we re-arrange it:

${4\choose1}+\cdots+{51\choose49}+{52\choose 48}+{53\choose47}$

It's the hockey stick identity!

All we're missing is the 3 choose 0 term. This can be solved easily:

${4\choose1}+\cdots+{51\choose49}+{52\choose48}+{53\choose47}={54\choose47}-{3\choose0}={54\choose3}-1$