Answer
$83160$
Work Step by Step
In "ABRACADABRA", there are 5 A's, 2 B's, 1 C, 1 D, and 2 R's, for a total of 11 letters. So we are trying to calculate the number of permutations of 11 letters, with the 5 A's indistinguishable from each other, 2 B's indistinguishable from each other, and 2 R's indistinguishable from each other. By Theorem 3 on page 428, the number of permutations is:
$\frac{11!}{5!2!2!}$
$=\frac{1*2*3*4*5*6*7*8*9*10*11}{1*2*3*4*5*1*2*1*2}$
$=\frac{6*7*8*9*10*11}{2*2}$
$=83160$