What's the number of arrangements of all 7 letters of the word "MINIMUM" in which the letters M are separated? I came across a question in a exerise booklet for Mathematical Olympiad for year 6 primary school students in Australia. The questions is as follows:
Find the number of arrangements of all 7 letters of the word "MINIMUM" in which the letters M are separated.
I calculated the number of arrangements as 360, but the provided keys says it's 720 instead. I'm not sure whether the key is right because the totoal number of arrangements when there are no restrictions is 420. I don't know why the key is 720. Thank you for your help.
 A: The correct answer should be 96 (of course, without duplicate arrangements). There are many ways to solve this problem.
Method 1. Start with arrangements of _ I _ N _ I _ U _ and place 3 Ms in 3 of 5 places. Then subtract arrangements where 2 I's are together
$$= \frac{4!}{2!} . \binom{5}{3} - 3!.\binom{4}{3}$$
Method 2. Start with arrangements of _ N _ U _ and place 2 I's in 2 of 3 places. This results in 4 letters and 5 adjacent places. Then place 3 M's in 3 of 5 places. Then add arrangements where I's are together. Place one M between two I's and 2 M's in 2 of  resulting 4 places.
$$= 2!.\binom{3}{2}.\binom{5}{3} + 2!.\binom{3}{1}.\binom{4}{2}$$
Method 3. Start with arrangements of _ N _ U _ and place 3 M's in 3 places. This results in 5 letters and 6 adjacent places. Then place 2 I's in 2 of 6 places. Then add arrangements where 2 M's are together. Place the other M in 1 of 2 remaining places. Then place one I between two M's and other I in resulting 5 places. The last step will be to add arrangement where all M's are together. Place 2 I's between 3 M's to separate them.
$$= 2!.\binom{3}{3}.\binom{6}{2} + 2!.\binom{3}{1}.\binom{2}{1}.\binom{5}{1} + 2!.\binom{3}{1}$$
