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help pls

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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?

Jun 25, 2020

#1
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Jun 25, 2020
#2
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well its incorrect

Jun 25, 2020
#3
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Have you tried anything at the very least? and what counting and probability topic is this?

gwenspooner85  Jun 25, 2020
edited by gwenspooner85  Jun 25, 2020
edited by gwenspooner85  Jun 25, 2020
#4
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yes ive tried doing it but its wrong

Jun 25, 2020
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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.

Jun 25, 2020
edited by amazingxin777  Jun 25, 2020
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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:

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

Jun 30, 2020
#7
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1.)

so there are 46 people voting for 4 people so there are 46*4=184 possible votes

2.)

it's the same exept we do 46*5=230

`These might NOT be the correct answer but they could be the correct answer but they might not be`
Jul 2, 2020