1860 United States presidential election in Vermont
|Elections in Vermont|
The 1860 United States presidential election in Vermont took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose five electors of the Electoral College, who voted for president and vice president.
With 75.86 percent of the popular vote, Lincoln's victory with in the state would be his strongest victory in terms of percentage in the popular vote.
|1860 United States presidential election in Vermont|
|Party||Candidate||Running mate||Popular vote||Electoral vote|
|Republican||Abraham Lincoln of Illinois||Hannibal Hamlin of Maine||33,808||75.86%||5||100%|
|Democratic||Stephen A. Douglas of Illinois||Herschel Vespasian Johnson of Georgia||8,649||19.41%||0||0.00%|
|Southern Democratic||John C. Breckinridge of Kentucky||Joseph Lane of Oregon||1,866||4.19%||0||0.00%|
|Constitutional Union||John Bell of Tennessee||Edward Everett of Massachusetts||217||0.49%||0||0.00%|
- "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1860 Presidential General Election Results - Vermont".
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