#### Answer

$831,600$

#### Work Step by Step

See Permutations of Duplicate Items, page 699.
The number of permutations of $n$ items, where $p$ items are identical, $q$ items are identical, $r$ items are identical, and so on, is
$\displaystyle \frac{n!}{p!q!r!\ldots}$
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TALLAHASSEE contains n=11 letters, of which there are
3 duplicate A's ... p=3,
2 duplicate L's ... q=2
2 duplicate S's .... r=2
2 duplicate E's ... s=2
a single T and a single H
..... in all n=11 (checking) ...
So, the total number of distinct permutations is
$\displaystyle \frac{11!}{3!2!2!2!}=$
$\dfrac{11\times 10\times 9\times 8\times 7\times(6)\times 5\times(4)\times 3\times(2)\times 1}{(6)\times(2\times 2)\times(2)}$
$=11\times 10\times 9\times 8\times 7\times 5\times 3$
$=831,600$