1880 United States presidential election in Indiana
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A presidential election was held in Indiana on November 2, 1880, as part of the 1880 United States presidential election. The Republican ticket of the U.S. representative from Ohio's 19th congressional district James A. Garfield and the chair of the New York Republican Party Chester A. Arthur defeated the Democratic ticket of the major general Winfield S. Hancock and the former U.S. representative from Indiana's 2nd congressional district William H. English. Garfield defeated Hancock in the national election with 214 electoral votes.[2]
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November 2, 1880
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| Turnout | 94.4%[1]  0.2 pp | |||||||||||||||||||||||||
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 County Results
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General election
Summary
Indiana chose 15 electors on a statewide general ticket. Nineteenth-century election laws required voters to elect each member of the Electoral College individually, rather than as a group. This sometimes resulted in small differences in the number of votes cast for electors pledged to the same presidential candidate, if some voters did not vote for all the electors nominated by a party.[3] This table reflects the statewide popular vote as calculated by Walter Dean Burnham in his influential study, Presidential Ballots, 1836â1892.[4] Svend Petersen finds six more votes for Garfield, one fewer for Hancock, and 80 fewer for Weaver in Indiana.[2]
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Republican | James A. Garfield Chester A. Arthur |
232,158 | 49.32 |  1.19 | |
| Democratic | Winfield Scott Hancock William Hayden English |
225,523 | 47.91 | 1.74 | |
| Greenback Labor | James B. Weaver Barzillai J. Chambers |
13,066 | 2.78 | 0.56 | |
| Total votes | 470,747 | 100.00 | |||
