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?

mathmathj28 Feb 25, 2020

#1**+1 **

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

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

.Rom Feb 25, 2020