I came across two questions which requires us to use the inclusion-exclusion principle to solve them. (That is method given by the course material) However, I thought that an alternate method might be possible, which worked for the first question, but did not work for the second. What am I missing in applying the same method to the second question?
First question- Find the number of ways all the letters of the word "MUNMUN" can be arranged if no two alike letters are together.
Since all the letters in the word repeat the same number of times, my approach was to first select any two alike letters, say, M. So now, there are three gaps between the letters where the next two alike letters can be placed (_ M _ M _).So we select any two gaps from the three available ones. Next, we have five gaps between the letters, out of which any two can be chosen, where the last pair of alike letters can be placed ( _ M _ N _ M _ N _). So, the answer would be $\binom{3}{2} \times \binom{5}{2} = 30$, which is the same answer you get by applying the inclusion-exclusion principle.
Second question- Find the number of ways all the letters of the word "HONOLULU" can be arranged if no two alike letters are together.
Though I do understand that we cannot apply the above method as it is, because H and N come in the word only once, can't we tweak the method to get an answer? I first reasoned that H and N can be arranged in two different ways, so, using the above analogy, I get $2 \times \binom{3}{2} \times \binom{5}{2} \times \binom{7}{2} = 1260$, which is different from the answer I am getting by the inclusion-exclusion principle $(2220)$
What am I missing here?