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.
 | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
 County Results
| ||||||||||||||||||||||||||
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] | |||||
|---|---|---|---|---|---|
| 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
editReferences
edit- ^ "Presidential Elections". Weekly Messenger. November 12, 1836.
- ^ "1836 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved August 4, 2012.
 | This Indiana elections-related article is a stub. You can help Wikipedia by expanding it. |