1840 United States presidential election in Indiana information
Election in Indiana
Main article: 1840 United States presidential election
1840 United States presidential election in Indiana
← 1836
October 30 - December 2, 1840
1844 →
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
65,302/65,280
51,604/51,696
Percentage
55.86%/55.79
44.14%/44.18%
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
Preview warning: Page using Template:Infobox election with unknown parameter "after_color"
Preview warning: Page using Template:Infobox election with unknown parameter "before_color"
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The 1840 United States presidential election in Indiana took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. 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%./11.61%
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