See main article: 1836 United States presidential election.
| Election Name: | 1836 United States presidential election in Indiana |
| Country: | Indiana |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1832 United States presidential election in Indiana |
| Previous Year: | 1832 |
| Next Election: | 1840 United States presidential election in Indiana |
| Next Year: | 1840 |
| Election Date: | November 3 – December 7, 1836 |
| Image1: | William Henry Harrison by James Reid Lambdin, 1835 crop.jpg |
| Nominee1: | William Henry Harrison |
| Party1: | Whig Party (United States) |
| Home State1: | Ohio |
| Running Mate1: | Francis Granger |
| Electoral Vote1: | 9 |
| Popular Vote1: | 41,281 |
| Percentage1: | 55.97% |
| Nominee2: | Martin Van Buren |
| Party2: | Democratic Party (United States) |
| Home State2: | New York |
| Running Mate2: | Richard Johnson |
| Electoral Vote2: | 0 |
| Popular Vote2: | 32,478 |
| Percentage2: | 44.03% |
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%.
| 1836 United States presidential election in Indiana[1] | |||||
|---|---|---|---|---|---|
| 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 | ||