1860 United States presidential election in New York
The 1860 United States presidential election in New York took place on November 6, 1860, as part of the 1860 United States presidential election. Voters chose 35 electors of the Electoral College, who voted for president and vice president.
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| Turnout | 95.5%[1]  5.6 pp | |||||||||||||||||||||||||
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| Elections in New York State |
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New York was won by Republican candidate Abraham Lincoln, who defeated the Democratic fusion ticket.
Lincoln won New York by a margin of 7.42%.
New York in the election was one of the 4 states that had a fusion ticket for the Democratic Party. The other three states were New Jersey, Pennsylvania and Rhode Island.
Results
| 1860 United States presidential election in New York[2] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Republican | Abraham Lincoln | 362,646 | 53.71% | 35 | |
| Fusion[lower-alpha 1] | Stephen A. Douglas / John Bell / John C. Breckinridge | 312,510 | 46.29% | 0 | |
| Totals | 675,156 | 100.0% | 35 | ||
Notes
- The slate of electors were pledged to 3 different candidates: 18 to Douglas, 10 to Bell, and 7 to Breckinridge.[3]
References
- Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- "1860 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 3 August 2012.
- Dubin, Michael J., United States Presidential Elections, 1788–1860: The Official Results by County and State, McFarland & Company, 2002, p. 187
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