jmc

There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations. Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple. There are now 813 = 243 permutations, but we have overcounted. Some multiples of 11 have a zero, and we must subtract a permutation for each. There are 110, 220, 330 ... 990, yielding 9 extra permutations Also, there are 209, 308, 407...902, yielding 8 more permutations. Now, just subtract these 17 from the total (243), getting $\boxed{226}$.