1840 United States presidential election in Indiana

Main article: 1840 United States presidential election

1840 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 John Tyler none Electoral vote 9 0 Popular vote 65,302 51,604 Percentage 55.86% 44.14%

County Results Harrison   50–60%   60–70%   70–80% Van Buren   50–60%   60–70%   70–80%   80–90% Unknown/No vote

President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

A presidential election was held in Indiana on November 2, 1840 as part of the 1840 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 the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Indiana by a margin of 11.72%.

Results

1840 United States presidential election in Indiana [2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Whig	William Henry Harrison of Ohio	John Tyler of Virginia	65,302	55.86%	9	100.00%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	51,604	44.14%	0	0.00%
Total				116,906	100.00%	9	100.00%

See also

• United States presidential elections in Indiana

References

1. Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
2. "1840 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved December 23, 2013.