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Director
Joined: 03 Jul 2003
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There are 3 candidates and 5 voters. In how many ways can [#permalink]
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06 Jan 2004, 15:18
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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 ?



Director
Joined: 28 Oct 2003
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Location: 55405

Quote: In how many ways can the votes be given?
does the identity of the vote matter?
In other words, say Bob got one vote does it matter if the vote Bob got was from Fran or Sally?



GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

1 3 * 5 = 15
2 2 (voters left) * 3 = 6
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Paul



Director
Joined: 03 Jul 2003
Posts: 652

I got this from another web site. So, I don't know the interpretation of
the question!



GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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
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Paul



Director
Joined: 14 Oct 2003
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Location: On Vacation at My Crawford, Texas Ranch

Re: Comb  voting [#permalink]
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06 Jan 2004, 16:52
kpadma wrote: 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*(himselfbut 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
i may be completely wrong.



CEO
Joined: 15 Aug 2003
Posts: 3454

Re: Comb  voting [#permalink]
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07 Jan 2004, 04:30
kpadma wrote: There are 3 candidates and 5 voters. In how many ways can the votes be given? Ans with explanation ? i smell permutations! 5 0 0..... 3 ways ( since 0 is not unique ) 4 1 0..... 6 ways ( 3! ways) 3 1 1..... 6 ways 2 2 1. ... 6 ways Number of ways = 21 ways Quote: If 3 candidates are included in 5 voters and they vote for themselves only, then how many ways of giving votes are possible?
3 candidates are included in 5 voters.
we got 2 voters and find out how many ways votes can be given to 3 candidates ( the other 3 votes have been assigned)
2 0 0 .... 3 ways
1 1 0 .... 6 ways
Total 9 ways



Director
Joined: 03 Jul 2003
Posts: 652

All I can say is it is a PERMUTATION problem and the
answers are 243 and 9.
Any brave souls want to reason how to get this answers



Director
Joined: 28 Oct 2003
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Location: 55405

243= 3^5
So the identity of of the voter matters!



Director
Joined: 03 Jul 2003
Posts: 652

stoolfi wrote: 243= 3^5
So the identity of of the voter matters!
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?



Director
Joined: 28 Oct 2003
Posts: 501
Location: 55405

Quote: Could you explain the second question? Yep. Three voters each vote for themselves, so just forget about them, ad there can only be one outcome as far as they are concerned. As for the others, two voters, three choices apiece 3^2. Quote: Whak kind of permutation is it? I just do the problems. I do not name names, here. One thing I could not understand is that how small samples such as 3 and 5 can give 243 permutations! that little carat (^) is one powerful thing! Quote: Could any one suggest any material in the web or books that specifically explains these kind of (2 or more dimentional) permutation problems?
AkamaiBrah, who is certainly a greater GMAT genius than I, recommends the Schaum guides.
http://www.amazon.com/exec/obidos/tg/de ... 9?v=glance
http://www.amazon.com/exec/obidos/tg/de ... ce&s=books
If you search used bookstores, I'm sure they can be had pretty cheaply.



SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

I thought it should be 5^3



SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

5^3 does not make sense. Three candidates are included in the voters. I read the q wrong.
I take it back



SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

Hi stoolfi your answer is not clear.
Let the candidates by C1,C2,C3 and voters be V1 and v2
C1 can get C1, C1V1, C1V2, C1V1V2 the same thing applies for each candidate.
I get 9 combinations = 3C2 + 3



Director
Joined: 28 Oct 2003
Posts: 501
Location: 55405

anand,
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 times3^5
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.
Three possible votes, 2 times 3^2
To illustrate:
1 2 3 1 1
1 2 3 1 2
1 2 3 1 3
1 2 3 2 1
1 2 3 2 2
1 2 3 2 3
1 2 3 3 1
1 2 3 3 2
1 2 3 3 3
Make sense yet?



SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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.



SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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.










