1860 United States presidential election in Indiana
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 thirteen representatives, or electors, to the Electoral College, who voted for president and vice president.
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| Elections in Indiana | ||||||
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Indiana was won by Illinois Representative Abraham Lincoln (R–Kentucky), running with Senator Hannibal Hamlin, with 51.09% of the popular vote, against Senator Stephen A. Douglas (D–Vermont), running with 41st Governor of Georgia Herschel V. Johnson, with 42.44% of the popular vote.
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Abraham Lincoln | 139,033 | 51.09% | |
| Democratic | Stephen A. Douglas | 115,509 | 42.44% | |
| Southern Democratic | John C. Breckinridge | 12,295 | 4.52% | |
| Constitutional Union | John Bell | 5,306 | 1.95% | |
| Total votes | 272,143 | 100% | ||
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References
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