A certain club has people, and members are running for president. Each club member either votes for one of the candidates, or can abstain from voting. How many different possible vote totals are there?
+38 Hi are there any numbers that go with this post? because with none it's unsolvable.
Omg sorry here's the new question:
A certain club has 50 people, and 4 members are running for president. Each club member - including the members running for president - votes for one of the 4 candidates. Candidates may vote for themselves or for a different candidate. How many different possible vote totals are there?
+38 Hey actually this question was solved
here View question - help pls (0calc.com)
Go to the 6th answer it explains the best.
+38 do you know combinations?
for this question you will use the combination formula which is
\(n!/r!(n-r)!\)
in this case your n is 50 and your r value is 4 because you have 50 options are you are choosing 4.
When plugged into the formula this gives you
\(50!/4!(50-4)!\)
simplified this is
\(50!/4!(46)!\)
can you solve from here?
+38 never mind I was given the wrong question to solve this one is slightly more complex but still doable haha
+2489 Two different questions are posted on this thread.
Question 1:
A certain club has [50] people, and [4] members are running for president. Each club member either votes for one of the candidates, or can abstain from voting. How many different possible vote totals are there?
Question 2:
A certain club has 50 people, and 4 members are running for president. Each club member - including the members running for president - votes for one of the 4 candidates. Candidates may vote for themselves or for a different candidate. How many different possible vote totals are there?
Solutions Below....
GA
--. .-
+2489
Solution for Q1:
(Because a member can abstain, this is counted as a vote for no candidate.)
Case one (1): All candidates receive at least one vote, including the no candidate, which means there is at least one abstention.
Partitions of 50 with a size of 5 = 2611
Case one (2): One candidate receives zero (0) votes; all others receive at least one vote.
Partitions of 50 with a size of 4 = 920
Case one (3): Two candidates receive zero (0) votes.
Partitions of 50 with a size of 3 = 208
Case one (4): Three candidates receive zero (0) votes.
Partitions of 50 with a size of 2 = 25
Case one (5): Four candidates receive zero (0) votes. This means everyone abstains, i.e. everyone votes for no candidate
Partitions of 50 with a size of 1 = 1
Total distribution of votes: 2611 + 920 + 208 + 25 + 1 = 3765
------------------
Solution for Q2:
(This is same as question 1, except the members cannot abstain.)
Case one (1): All candidates receive at least one vote
Partitions of 50 with a size of 4 = 920
Case one (2): One candidate receives zero (0) votes; all others receive at least one vote.
Partitions of 50 with a size of 3 = 208
Case one (3): Two candidates receive zero (0) votes.
Partitions of 50 with a size of 2 = 25
Case one (4): Three candidates receive zero (0) votes. One candidate receives all the votes.
Partitions of 50 with a size of 1 = 1
Total distribution of votes: 920 + 208 + 25 + 1 = 1154
GA
--. .-
