1824 United States presidential election in Indiana

Main article: 1824 United States presidential election

1824 United States presidential election in Indiana

←  1820 October 26 – December 2, 1824 1828  →

Nominee Andrew Jackson Henry Clay John Quincy Adams Party Democratic-Republican Democratic-Republican Democratic-Republican Home state Tennessee Kentucky Massachusetts Running mate John C. Calhoun Nathan Sanford John C. Calhoun Electoral vote 5 0 0 Popular vote 7,343 5,315 3,095 Percentage 46.61% 33.74% 19.65%

County Results Jackson   40–50%   50–60%   60–70%   70–80% Clay   30–40%   40–50%   50–60%   60–70%   80–90%   90–100% Adams   40–50%   50–60% Unknown/No vote

President before election James Monroe Democratic-Republican Elected President John Quincy Adams Democratic-Republican

The 1824 United States presidential election in Indiana took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Indiana voted for Andrew Jackson over Henry Clay and John Quincy Adams. Jackson won Indiana by a margin of 12.87%.

Results

1824 United States presidential election in Indiana [1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic-Republican	Andrew Jackson	7,343	46.61%	5
	Democratic-Republican	Henry Clay	5,315	33.74%	0
	Democratic-Republican	John Quincy Adams	3,095	19.65%	0
Totals			15,753	100.0%	5

See also

• United States presidential elections in Indiana

References

1. "1824 Presidential General Election Results - Indiana". U.S. Election Atlas. Retrieved February 27, 2013.