1840 United States presidential election in New York
(Redirected from United States presidential election in New York, 1840)
The 1840 United States presidential election in New York took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 42 representatives, or electors to the Electoral College, who voted for President and Vice President.
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| Turnout | 91.9%[1]  21.4 pp | |||||||||||||||||||||||||
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 County Results
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New York voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New York by a narrow margin of 3.00%.
Results
edit| 1840 United States presidential election in New York[2] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 226,001 | 51.18% | 42 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 212,733 | 48.18% | 0 | 0.00% | ||
| Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 2,809 | 0.64% | 0 | 0.00% | ||
| Total | 441,543 | 100.00% | 42 | 100.00% | ||||
See also
editReferences
edit- ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- ^ "1840 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved December 23, 2013.
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