Discrete Math problem combinations and restrictions I am trying to get my head around an idea but I can't seem to get it to work.
Imagine you have the word "MAMMAL"
Lets see I wanted to figure out how many ways I could rearrange the letters. Well that is easy. It is simply
6!/3!2! = 60 possibilities
But what if I added a restriction to it such that all M's must be together.
Then I could consider all M's as one letter and it would be 4!/2! = 12 possibilities.
Again I understand that. But what if the restriction was there needs to be a minimum of 2 M's together at all times.
If I were to do the same process as the previous example, I would combine 2 M's into one. Which then would become 5!/2! = 60.
This seems to be a wrong answer because it is the same as my first calculation of finding all possibilities without any restrictions. Can anyone please explain to me as to how I need to approach the last problem of finding number of combinations where at least 2 M's are always together? Thanks!

Consider the complimentary event, which is no Ms are together, i.e. all Ms are separate.
3 Ms need to be put in 6 places. There are only 4 choices, M _ M _ M _,  _ M _ M _M, M _ M _ _ M and M _ _ M _ M.
In remaining 3 places, you need to put 3 letters, out of which 2 are different. Therefore, there are $4\times\dfrac{3!}{2!} = 12$ such words ($4$ for $4$ different arrangements described above).
Subtract this from total words $60$, and you'll get your answer as $48$.
