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Say I want to arrange the letters of the word ISOMORPHISM, such that no two vowels are next to each other but the vowels are in alphabetical order.

What I'd do is firstly consider the consonants SMRPHSM, which in total have 7 letters and thus 8 'gaps'.

| S | M | R | P | H | S | M |

In these 8 gaps, I place 4 vowels, so $\binom{8}{4}$

I do not arrange these vowels because there is only one such permutation where they are in alphabetical order.

Now, I arrange the consonants amongst themselves, so $\frac{7!}{2!2!}$ as there are two M's and two S's.

So would the correct solution be $\binom{8}{4} \times \frac{7!}{2!2!}$ ?

The reason why I am asking is because I'm not sure if I should divide by 2!2! again, as there are two I's and two O's as well.

My gut feeling is telling me 'No, you do not need to do so again because you did not permute the vowels', but I just want to confirm with you.

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up vote 2 down vote accepted

Your analysis is correct, and quite clearly put.

Once you have chosen the "gaps" the location of the I's and O's is completely determined because of the "alphabetical order" condition. Since you have done no double-counting, there is no need to adjust by a division.

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