1860 United States presidential election in Indiana explained

See main article: 1860 United States presidential election.

Election Name:1860 United States presidential election in Indiana
Country:Indiana
Flag Year:1846
Type:presidential
Ongoing:no
Previous Election:1856 United States presidential election in Indiana
Previous Year:1856
Next Election:1864 United States presidential election in Indiana
Next Year:1864
Election Date:November 6, 1860
Image1:Abraham Lincoln by Alexander Hesler, 1860-restored (cropped).png
Nominee1:Abraham Lincoln
Party1:Republican Party (United States)
Home State1:Illinois
Running Mate1:Hannibal Hamlin
Electoral Vote1:13
Popular Vote1:139,033
Percentage1:51.09%
Nominee2:Stephen A. Douglas
Party2:Democratic Party (United States)
Home State2:Illinois
Running Mate2:Herschel V. Johnson
Electoral Vote2:0
Popular Vote2:115,509
Percentage2:42.44%
President
Before Election:James Buchanan
Before Party:Democratic Party (United States)
After Election:Abraham Lincoln
After Party:Republican Party (United States)
Map Size:300px

The 1860 United States presidential election in Indiana took place on November 6, 1860, as part of the 1860 United States presidential election. Indiana voters chose 13 representatives, or electors, to the Electoral College, who voted for president and vice president.

Indiana was won by Republican nominee Representative Abraham Lincoln of Illinois and his running mate Senator Hannibal Hamlin of Maine. They defeated the Democratic nominee Senator Stephen A. Douglas of Illinois] and his running mate 41st Governor of Georgia Herschel V. Johnson. Lincoln won the state by a margin of 8.65%.

See also