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An election was held using the conventional Borda count method. There were four candidates \((A, B, C,\) and \(D)\) and 110 voters. When the points were tallied (using 4 points for first, 3 points for second, 2 points for third, and 1 point for fourth), \(A\) had 320 points, \(B\) had 290 points, and \(C\) had 180 points. Find how many points \(D\) had and give the ranking of the candidates.

Imagine that in the voting for the American League Cy Young Award ( 7 points for first place, 4 points for second, 3 points for third, 2 points for fourth, and 1 point for fifth) there were five candidates \((A, B, C, D,\) and \(E)\) and 50 voters. When the points were tallied \(A\) had 152 points, \(B\) had 133 points, \(C\) had 191 points, and \(\mathrm{D}\) had 175 points. Find how many points \(E\) had and give the ranking of the candidates.

An election with five candidates \((A, B, C, D,\) and \(E)\) is decided using the method of pairwise comparisons. If \(B\) loses two pairwise comparisons, \(C\) loses one, \(D\) loses one and ties one, and \(E\) loses two and ties one, (a) find how many pairwise comparisons \(A\) loses. (b) find the winner of the election.

An election with six candidates \((A, B, C, D, E,\) and \(F)\) is decided using the method of pairwise comparisons. If \(A\) loses four pairwise comparisons, \(B\) and \(C\) both lose three, \(D\) loses one and ties one, and \(E\) loses two and ties one, (a) find how many pairwise comparisons \(F\) loses. (b) find the winner of the election.

Explain why the method of pairwise comparisons satisfies the Condorcet criterion.

Explain why the plurality method satisfies the monotonicity criterion.

Explain why the Borda count method satisfies the monotonicity criterion.

Explain why the method of pairwise comparisons satisfies the monotonicity criterion.

The following simple variation of the conventional Borda count method is sometimes used: last place is worth 0 points, second to last is worth 1 point,..., first place is worth \(N-1\) points (where \(N\) is the number of candidates). Explain why this variation is equivalent to the conventional Borda count described in this chapter (i.e., it produces exactly the same winner and the same ranking of the candidates).

The AP college football poll is a ranking of the top 25 college football teams in the country. The voters in the AP poll are a group of sportswriters and broadcasters chosen from across the country. The top 25 teams are ranked using a conventional Borda count: a first-place vote is worth 25 points, a second- place vote is worth 24 points, a third-place vote is worth 23 points, and so on. A last-place vote is worth 1 point. Table \(1-44\) shows the ranking and total points for each of the top three teams at the end of the 2006 regular season. (The remaining 22 teams are not shown here because they are irrelevant to this exercise.) $$\begin{array}{|l|c|}\hline {\text { Team }} & \text { Points } \\\\\hline 1 . \text { Ohio State } & 1625 \\\\\hline \text { 2. Florida } & 1529 \\\\\hline \text { 3. Michigan } & 1526 \\\\\hline\end{array}$$ (a) Given that Ohio State was the unanimous first-place choice of all the voters, find the number of voters that participated in the poll. (b) Given that all the voters had Florida in either second or third place, find the number of second-place and the number of third-place votes for Florida. (c) Given that all the voters had Michigan in either second or third place, find the number of second-place and the number of third-place votes for Michigan.