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# a) In a club with 50 members, four of the members are running for president. Each of the 50 members votes for one of the four candidates.

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a) In a club with 50 members, four of the members are running for president. Each of the 50 members votes for one of the four candidates. How many different vote counts are possible?

b) In a club with 50 members, four of the members are running for president. Each of the 50 members either votes for one of the four candidates, or can abstain. How many different vote counts are possible?

Feb 25, 2020

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$$\text{a) is equivalent to putting 50 identical balls into 4 distinct boxes}\\ \text{This is the well known stars and bars problem}\\ N=\dbinom{50+4-1}{4-1} = \dbinom{53}{3} = 23426$$

$$\text{b) is essentially the same as (a) but now there is a 5th box}\\ N=\dbinom{50+5-1}{5-1} = 316251$$

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Feb 25, 2020
edited by Rom  Feb 25, 2020
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Thanks so much!!

mathmathj28  Feb 25, 2020