A certain club has $50$ people, and $4$ members are running for president. Each club member either votes for one of the $4$ candidates, or can abstain from voting. How many different possible vote totals are there?

Guest Jun 25, 2020

#1**+1 **

This has been posted before: https://web2.0calc.com/questions/help-please_25835

gwenspooner85 Jun 25, 2020

#2

#3**+1 **

Have you tried anything at the very least? and what counting and probability topic is this?

gwenspooner85
Jun 25, 2020

#5**+1 **

If 4 members are running for president, then only 46 people can vote (unless you can vote for yourself???). Each club member (of the 46) has 5 choices: don't vote, or vote for one of the four club members. That means it is \(\boxed {5^{46}}\). An expert in C&P may need to check my answer.

amazingxin777 Jun 25, 2020

#6**0 **

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:

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

.AmyMath03 Jun 30, 2020