A panel consisting of five experts takes the interview of the four short listed candidates. If one of the selectors has a personal grudge against a particular candidate, and would never vote for him/her, how to find the total number of ways the votes can be cast?
The biased interview
-
3This is not very clear: do we only care about vote totals (is there a secret ballot among the experts and we only care about how many votes each candidate got), or is who voted for whom important (we care about which expert voted for which candidate)? – 2011-11-12
2 Answers
I’m assuming that the number of ways the votes can be cast treats the experts as well as the candidates as distinguishable, so that there are $5$ different ways in which Candidate A can receive $4$ votes and Candidate B only $1$ vote (assuming that neither A nor B is the candidate against whom one expert has a grudge). If only the vote tally matters, then you want Arturo’s answer.
The selector with the grudge can cast his vote in $3$ ways. Each of the other four experts can cast his vote in $4$ ways. The total number of ways in which they can cast their votes is therefore ... ?
-
0How the selector with the grudge caste is vote in 4 ways? there is 4 candidates only isn't? – 2011-11-12
-
0@MaX: Sorry: both of those numbers are too large by $1$. For some reason I was thinking that there were $5$ candidates. I’ll fix it in a moment. – 2011-11-12
-
1therefore $3\times 4^4=768$ – 2011-11-12
-
1@MaX: Yep. $\qquad$ – 2011-11-13
-
0The official solution just came out,and this is the interpretation they followed. – 2011-11-13
I assume you only want to know the possible tally totals, with candidates distinguishable but votes indistinguishable. That is, votes are secret so you can only know the totals. If the committee votes in the open and we care about who voted for whom, then you want Brian Scott's answer.
If there are no restrictions, then you are just making 5 selections among 4 possibilities, allowing repetitions and where order does not matter. This is equivalent to rolling five 4-sided dice, or to a stars-and-bars problem. The number of possibilities is $$\binom{5+4-1}{5} = \binom{8}{5} = 56.$$ Of all the possible vote tallies, the only one that cannot occur when one panelist has a grudge against, say, Candidate 1, is a unanimous vote for candidate one. Every other vote tally can occur. So the total number of possible vote tallies is therefore $56-1=55$.
If we only know that some interviewer has a grudge against some interviewee, the number of possible vote tallies is still $55$, though we don't know who cannot get a unanimous vote.
-
0Given MaX’s previous questions, your interpretation seems likelier than mine. – 2011-11-12
-
0@Brian M. Scott:My initial approach was exactly like yours (in the deleted answer),but as Arturo stated in the comment the question is not clear, and this answer is whole new approach of looking at the problem. – 2011-11-12
-
0@Brian:I agree Arturo, you may like to un-delete your answer, as I can't think of any cogent argument why that approach should not work here. – 2011-11-12
-
2@MaX, Arturo: Okay; done. – 2011-11-12