1836 United States presidential election in Indiana
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| Elections in Indiana |
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A presidential election was held in Indiana on November 7, 1836 as part of the 1836 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 Whig candidate William Henry Harrison over the Democratic candidate, Martin Van Buren. Harrison won Indiana by a margin of 11.94%.
Results
[edit]| 1836 United States presidential election in Indiana[2] | |||||
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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 | ||
See also
[edit]References
[edit]- ^ "Presidential Elections". Weekly Messenger. November 12, 1836.
- ^ "1836 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved August 4, 2012.
State and district results of the 1836 United States presidential election | ||
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