1808 United States presidential election in Georgia
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Elections in Georgia |
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The 1808 United States presidential election in Georgia took place between 4 November and 7 December 1808, as part of the 1808 United States presidential election. The state legislature chose six representatives, or electors to the Electoral College, who voted for President and Vice President.
Georgia cast six electoral votes for the Democratic-Republican candidate James Madison over the Federalist candidate Charles C. Pinckney. The electoral votes for Vice president were cast for Madison's running mate George Clinton from New York. These electors were elected by the Georgia General Assembly, the state legislature, rather than by popular vote.[1]
Results
1808 United States presidential election in Georgia[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | James Madison | – | – | 6 | |
Federalist | Charles C. Pinckney | – | – | 0 | |
Totals | – | – | 6 |
See also
References
- ^ "1808 Presidential General Election Results". U.S. Election Atlas. Retrieved July 10, 2023.
- ^ "1808 Presidential Election". 270towin.com. Retrieved July 10, 2023.
Categories:
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- 1808 United States presidential election by state
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- 1808 Georgia (U.S. state) elections
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- November 1808 events
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- 1808 in Georgia (U.S. state)