1836 United States presidential election in Indiana

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.

Main article: 1836 United States presidential election
1836 United States presidential election in Indiana
November 3 – December 7, 1836
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate Francis Granger Richard Johnson
Electoral vote 9 0
Popular vote 41,281 32,478
Percentage 55.97% 44.03%
Elections in Indiana
Federal government
U.S. President
U.S. Senate
U.S. House of Representatives
State government
State elections
Gubernatorial elections
State Senate elections
State House of Representatives elections
Evansville
Mayoral Elections
Fort Wayne
Mayoral Elections
Gary
Mayoral Elections
Indianapolis
Mayoral Elections
Before Unigov
Unigov-era
City-County Council Elections
South Bend
Mayoral Elections

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]
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

References
  1. "1836 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved 4 August 2012.


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