TITLE: How many ways can you arrange all letters in the word "management" so that at least one "m" is adjacent to one "a"?
QUESTION [5 upvotes]: This permutation question appears harder than it seems.
I first to combine a set of "m" and "a" together. Then subtract all cases where "ma" and the other "ma", or "am" and the other "am" as they can switch places. Then subtract the "ama", "mam" cases.
Eventually, get lost in the numerous duplicate cases, can someone please help me with this.
Thanks in advance.

REPLY [2 votes]: Here is another way to compute the complement, i.e. how many ways there are to arrange the letters with no MA or AM.
First, distribute the other letters EENNGTT; as noted in the other answer, there are $\dfrac{10!}{2!2!2!2!} = 180$ ways to do this.
Write them down and put slots where the remaining M's and A's could go, like:
_E_E_N_N_G_T_
So there are in principle $7$ slots for the remaining letters. Now we either

choose four of the slots, and then distribute the letters in them (automatically no MA or AM): $\binom{7}{4} \cdot \dfrac{4!}{2!2!} = 210$ options, or
choose three of the slots, then choose one of the three chosen one for a double letter, then choose whether we want a double M or double A (as MA and AM are forbidden): $\binom{7}{3} \cdot 3 \cdot 2 = 210$, or
choose two of the slots; if one contained three or four of our MAMA, we could not avoid an adjacent pair, so both have to have a two letters in them, i.e. one of them will have to have AA and the other MM in it; choose which one: $\binom{7}{2} \cdot 2 = 42$
if we chose just one slot for the AAMM, we could not avoid adjacent M and A, so that's it.

In total, we have $210+210+42=462$ options. Multiply with $180$ for the distribution of the other letters, gives
$$462 \cdot 180 = 83'160$$
ways to reorder the letters without MA or AM. Since the total number of arrangements is $\dfrac{10!}{2!2!2!2!} = 226'800$, the answer to your question is
$$226'800-83'160=143'640.$$