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There are 3 candidates and 5 voters. In how many ways can [#permalink]
06 Jan 2004, 14:18

There are 3 candidates and 5 voters. In how many ways can the votes be given? If 3 candidates are included in 5 voters and they vote for themselves only, then how many ways of giving votes are possible?

I don't know if you need this kpadma but here it is
Let's call voters ( A B C D E )
Let's call candidates ( 1 2 3 )

Question 1: Voter A can vote for 3 diff. candidates. So you have 3 possibilities. All other candidates can also vote for 3 diff. candidates. Therefore 3*5 = 15

Question 2: Each candidates will vote for themselves, therefore, there will only be 2 voters left. Let's call those left A and B ( C D E being the candidates who voted for themselves already ). A and B can each choose 3 candidates to vote fore. Thus 2*3 = 6 _________________

There are 3 candidates and 5 voters. In how many ways can the votes be given? If 3 candidates are included in 5 voters and they vote for themselves only, then how many ways of giving votes are possible?

Ans with explanation ?

maybe i'm not getting something here but I got something really different.

this is how i interpreted the question:

how many ways can 5 voters vote for 3 candidates?

Candidates X Y Z

X can get (6, 5, 4, 3, 2, 1*(himself-but i don't think this matters)

Thus

All 5 voters voting for Candidate X = 5C5

4 of 5 voters voting for candidate x =5C4

3 of 5 voters ditto = 5C3

2 of 5 voters ditto = 5C2

1 of 5 voters ditto = 5C1

= (1*5*10*10*5)*3 (could be x,y, or z) = 7,500 ways

Stoolfi
You are a genius!
Could you explain the second question?
Whak kind of permutation is it?
One thing I could not understand is that how small samples
such as 3 and 5 can give 243 permutations!

Could any one suggest any material in the web or books
that specifically explains these kind of (2 or more dimentional)
permutation problems?

your post is unclear to me, but I'll try again to explain.

In the first case,

There are 3 candidates and 5 voters. In how many ways can the votes be given?

we have five voters who can make three different choices

Voter one can pick A, B, or C.
Voter two can pick A, B, or C.
Voter three can pick A, B, or C.
Voter four can pick A, B, or C.
Voter five can pick A, B, or C.

3 possible votes, five times------3^5

Second question:

If 3 candidates are included in 5 voters and they vote for themselves only, then how many ways of giving votes are possible?

Voter 1 is candidate 1. He votes 1.
Voter 2 is candidate 2. He votes 2.
Voter 3 is candidate 3. He votes 3.

Voter 4 is not a candidate. He votes 1, 2, or 3.
Voter 5 is not a candidate. He votes 1, 2, or 3.

Man I completely screwedup. I did get 9 for the second question
For the first question I did a stupid thing.

This is what it is

I thought c1 will pick v1 to v5 so I got 5^3. How can a candidate pick voters? It actually otherway
v1 will pick either c1,c2,c3.
So the answer has to be 3^5 because each voter can pick any tree and there are 5 voters.

yesterdays beer is still working on me. I need to go and take a swim.

If another condition is added saying a voter cannot vote for two different candidates at the same time then 3^5 wont be the answer. I hope you will agree.