1836 United States presidential election in Indiana

1836 United States presidential election in Indiana

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Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate Francis Granger Richard M. Johnson Electoral vote 9 0 Popular vote 41,281 32,478 Percentage 55.97% 44.03%

County Results Harrison   50–60%   60–70%   70–80% Van Buren   50–60%   60–70%   70–80% Unknown/No vote

Main article: 1836 United States presidential election

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

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

• United States presidential elections in Indiana

References

1. "Presidential Elections". Weekly Messenger. November 12, 1836.
2. "1836 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved August 4, 2012.