1836 United States presidential election in Indiana
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| Elections in Indiana |
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The 1836 United States presidential election in Indiana took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for President and Vice President.
Indiana voted for Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Indiana by a margin of 11.94%.
Results
| 1836 United States presidential election in Indiana[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Whig | William Henry Harrison | 41,281 | 55.97% | 9 | |
| Democratic | Martin Van Buren | 32,478 | 44.03% | 0 | |
| Totals | 73,759 | 100.0% | 9 | ||
References
- ^ "1836 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved 4 August 2012.
State and district results of the 1836 United States presidential election | ||
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