I've been stuck on these for a while. Please guide me through all the steps because I actually want to understand this. I've got an exam coming up.
Consider the letters in the word "MATHEMATICS". In how many ways can these 11 letters be ordered so that:
(i) The two M's are next to each other.
(ii) The two M's are next to each other but the two A's are not.
(a) Treat the two $M$'s as a single unit. So we have now 10 letters. We permute these in $\frac{10!}{2! * 2!}$ ways. Since the $M$'s are identical, we don't have to permute the order in which the two $M$'s appear.
(b) This is an inclusion exclusion problem. You have from (a) the number of ways for the two $M$'s to appear together. Now group the two $A$'s together. So there are $9!$ ways of arranging the letters so that both $M$'s and both $A$'s are together. So subtract that out from your original answer: $\frac{10!}{2! * 2!} - \frac{9!}{2!}$.