1832 United States presidential election in Indiana

Summary

The 1832 United States presidential election in Indiana took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for President and Vice President.

1832 United States presidential election in Indiana

← 1828 November 2 – December 5, 1832 1836 →
 
Nominee Andrew Jackson Henry Clay
Party Democratic National Republican
Home state Tennessee Kentucky
Running mate Martin Van Buren John Sergeant
Electoral vote 9 0
Popular vote 31,551/31,652 15,472/25,473
Percentage 67.10%/55.38 32.90%/44.57

Indiana voted for the Democratic Party candidate, Andrew Jackson, over the National Republican candidate, Henry Clay. Jackson won Indiana by a margin of 10.81%/34.20%.

As of 2020, this remains the strongest performance by a Democrat in Indiana.[1](If 34.20% is correct than yes)

Results edit

1832 United States presidential election in Indiana[2]
Party Candidate Votes Percentage Electoral votes
Democratic Andrew Jackson (incumbent) 31,551 67.10% 9
National Republican Henry Clay 15,472 32.90% 0
Totals 47,023 100.0% 9

See also edit

References edit

  1. ^ "Dave Leip's Atlas of U.S. Presidential Elections". uselectionatlas.org. Retrieved January 10, 2023.
  2. ^ "1832 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved April 12, 2013.