1836 United States presidential election in Indiana

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United States presidential election in Indiana, 1836

← 1832 November 3 – December 7, 1836 1840 →
  William Henry Harrison by James Reid Lambdin, 1835 crop.jpg Van Buren.jpg
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%

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[edit]

United States presidential election in Indiana, 1836[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

References[edit]

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