Since the words start with "mi", we only need to find the distinguishable
permutations of "scellaneous" for "mi" can be placed in front of each of them.

If we could tell the two s's apart, the two e's apart, the two l's apart, the
answer would be 11!, but since we cannot tell them apart, we must divide by the
factorial of each of the numbers of the letters that we cannot tell apart.  So the
answer is:



Edwin