**QUESTION**

Text

Image

HELP IN ALL PLEASE

For an election with four candidates (A, B, C, D) consider the following preference table: \begin{tabular}{|l|c|c|c|c|} \hline $\begin{array}{l}\text { NUMBER OF } \\ \text { VOTERS }\end{array}$ & 6 & 3 & 5 & 8 \\ \hline $1^{\text {st }}$ Choice & D & D & A & C \\ \hline $2^{\text {nd }}$ Choice & B & A & C & A \\ \hline $3^{\text {rd }}$ Choice & A & B & B & D \\ \hline $4^{\text {th }}$ Choice & C & C & D & B \\ \hline \end{tabular} Using the Borda count method, which candidate wins the election? B C D A

Given the following preference table, the Condorcet winner is \begin{tabular}{|c|c|c|c|c|c|} \hline RANKING & 15 & 12 & 8 & 11 & 9 \\ \hline 1st & $\mathrm{Q}$ & $\mathrm{T}$ & $\mathrm{R}$ & $\mathrm{S}$ & $\mathrm{Q}$ \\ \hline 2nd & $\mathrm{S}$ & $\mathrm{Q}$ & $\mathrm{S}$ & $\mathrm{R}$ & $\mathrm{T}$ \\ \hline 3rd & $\mathrm{R}$ & $\mathrm{R}$ & $\mathrm{Q}$ & $\mathrm{T}$ & $\mathrm{S}$ \\ \hline 4th & $\mathrm{T}$ & $\mathrm{S}$ & $\mathrm{T}$ & $\mathrm{Q}$ & $\mathrm{R}$ \\ \hline \end{tabular} s Q R $\mathrm{T}$

Arrow's Impossibility Theorem implies that there is a voting method that satisfies all of the fairness criteria. none of these that in every election, each of the voting methods must produce a different winner that it is impossible to have a voting method that satisfies all of the fairness criteria.

Given the following preference table, the winner using the Hare method is \begin{tabular}{|c|c|c|c|c|c|} \hline RANKING & 15 & 12 & 8 & 11 & 9 \\ \hline 1st & Q & T & R & S & Q \\ \hline 2nd & S & Q & S & R & T \\ \hline 3rd & R & R & Q & T & S \\ \hline 4th & T & S & T & Q & R \\ \hline \end{tabular} R S $\mathrm{T}$ Q

For an election with four candidates (A, B, C, D) consider the following preference table: \begin{tabular}{|l|c|c|c|c|} \hline $\begin{array}{l}\text { NUMBER OF } \\ \text { VOTERS }\end{array}$ & 6 & 3 & 5 & 8 \\ \hline $1^{\text {st }}$ Choice & D & D & A & C \\ \hline $2^{\text {nd }}$ Choice & B & A & C & A \\ \hline $3^{\text {rd }}$ Choice & A & B & B & D \\ \hline $4^{\text {th }}$ Choice & C & C & D & B \\ \hline \end{tabular} Using the Condorcet method, which candidate wins the election? B C D A

For an election with four candidates (A, B, C, D) consider the following preference table: \begin{tabular}{|l|c|c|c|c|} \hline $\begin{array}{l}\text { NUMBER OF } \\ \text { VOTERS }\end{array}$ & 6 & 3 & 5 & 8 \\ \hline $1^{\text {st }}$ Choice & D & D & A & C \\ \hline $2^{\text {nd }}$ Choice & B & A & C & A \\ \hline $3^{\text {rd }}$ Choice & A & B & B & D \\ \hline $4^{\text {th }}$ Choice & C & C & D & B \\ \hline \end{tabular} Using the Borda count method, which candidate wins the election? B C D A

Given the following preference table, the Condorcet winner is \begin{tabular}{|c|c|c|c|c|c|} \hline RANKING & 15 & 12 & 8 & 11 & 9 \\ \hline 1st & $\mathrm{Q}$ & $\mathrm{T}$ & $\mathrm{R}$ & $\mathrm{S}$ & $\mathrm{Q}$ \\ \hline 2nd & $\mathrm{S}$ & $\mathrm{Q}$ & $\mathrm{S}$ & $\mathrm{R}$ & $\mathrm{T}$ \\ \hline 3rd & $\mathrm{R}$ & $\mathrm{R}$ & $\mathrm{Q}$ & $\mathrm{T}$ & $\mathrm{S}$ \\ \hline 4th & $\mathrm{T}$ & $\mathrm{S}$ & $\mathrm{T}$ & $\mathrm{Q}$ & $\mathrm{R}$ \\ \hline \end{tabular} s Q R $\mathrm{T}$

Arrow's Impossibility Theorem implies that there is a voting method that satisfies all of the fairness criteria. none of these that in every election, each of the voting methods must produce a different winner that it is impossible to have a voting method that satisfies all of the fairness criteria.

Given the following preference table, the winner using the Hare method is \begin{tabular}{|c|c|c|c|c|c|} \hline RANKING & 15 & 12 & 8 & 11 & 9 \\ \hline 1st & Q & T & R & S & Q \\ \hline 2nd & S & Q & S & R & T \\ \hline 3rd & R & R & Q & T & S \\ \hline 4th & T & S & T & Q & R \\ \hline \end{tabular} R S $\mathrm{T}$ Q

For an election with four candidates (A, B, C, D) consider the following preference table: \begin{tabular}{|l|c|c|c|c|} \hline $\begin{array}{l}\text { NUMBER OF } \\ \text { VOTERS }\end{array}$ & 6 & 3 & 5 & 8 \\ \hline $1^{\text {st }}$ Choice & D & D & A & C \\ \hline $2^{\text {nd }}$ Choice & B & A & C & A \\ \hline $3^{\text {rd }}$ Choice & A & B & B & D \\ \hline $4^{\text {th }}$ Choice & C & C & D & B \\ \hline \end{tabular} Using the Condorcet method, which candidate wins the election? B C D A