1840 United States presidential election in Indiana
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County Results
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| Elections in Indiana |
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A presidential election was held in Indiana on November 2, 1840 as part of the 1840 United States presidential election.[1] Voters chose nine representatives, or electors to the Electoral College, who voted for President and Vice President.
Indiana voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Indiana by a margin of 11.72%.
Results
| 1840 United States presidential election in Indiana[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 65,302 | 55.86% | 9 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 51,604 | 44.14% | 0 | 0.00% | ||
| Total | 116,906 | 100.00% | 9 | 100.00% | ||||
See also
References
- ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
- ^ "1840 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved December 23, 2013.