Arrangement of letters in the word $MINCEMEAT$ with condition

Question

Find the number of different ways in which all $9$ letters of the word $MINCEMEAT$ can be arranged if NO vowel $(A, E, I)$ is next to another vowel. Can anybody help me in this question?

Answer 1

Your permutations will look something like this - * A * E * I * E *, where at least one consonant must be in the middle three *s. First, permute A,E,I. That's $4\times 3\times 2/2=12$ ways. Now, you have 5 consonants left, two of which are M. Let's say for now the two M's are distinct.

Then you choose three of the five consonants to fill the middle three *'s. That's ${5\choose 3}=10$ ways. But then you can permute them in 6 ways, so that's 60 ways.

Now 2 consonants left and your word is looking like - * A * _ * E * _ * I * _ * E * where the three _'s in between have been filled and the *'s are available places. There are 8 *'s, so $8 \times 8$ places for the two.

Then those two consonants need to be permuted as well, so 2 ways for that. So far, the total is $12\times 60 \times 64 \times 2$.

But we assumed the two M's were distinct. So divide by 2. We end up with $12 \times 60 \times 64 = 46080$ ways.

EDIT: As people in the comments indicated, there is some double counting here. I'll call the number I got above an upper-bound.

Answer 2

The previous answer has been deleted, I presume that because it was not quite correct. But it was very close to being correct and was a very clever attempt and I will follow that! There are 9 letters, 5 consonants, and 4 vowels. In order that no two vowels be together the "form", taking "C" to represent a consonant and "V" to represent a vowel, the vowels and four of the consonants must be of the form CVCVCVCV or VCVCVCVC. We can then put the fifth C at any of the 5 places beside a V. That is, there are 2*5= 10 such combinations. The 4 vowels are I, E, E, A, four vowels, two of them the same. There are $\frac{4!}{2!}= 12$ ways to order them. The 5 consonants are M, N, M, C, T, five consonants but, again, two the same. There are $\frac{5!}{2!}= 60$ ways to order them. That gives a total of 10(12)(60)= 7200 ways to do this.

Answer 3

My previous answer was an over-estimation and @user247327 is an under-estimate. I want to give this another shot and hopefully, this attempt will stand up to peer review. This one does lie between @user247327 and my old answer, but closer to @user247327. As before, we know that the words will look like -

                                         * A * E * I * E * 

And as before, there are 12 permutations for the vowels. Now, we fill up the consonants into the positions given by the five stars. The question is how many consonants each star gets. We know that the middle three stars can't get less than one consonant.

Here are the possibilities:

                                           1 1 1 1 1 _(1)
                                           2 1 1 1 0 _(2)
                                           0 1 1 1 2 _(3)
                                           1 2 1 1 0 _(4) X3
                                           0 2 1 1 1 _(5) X3
                                           0 2 2 1 0 _(6) X3
                                           0 3 1 1 0 _(7) X3

We repeat configuration (4) three times because the 2 can be at any of the middle three stars. So, the total configurations for the consonants are 15.

In each configuration, we can permute the consonants in $\frac{5!}{2!}$ ways (because 2 M's).

This would make the total configurations: $12\times 15 \times 5 \times 4 \times 3 = 10800$.

Thanks to @N.F. Taussig for reviewing and pointing out corrections. Also, the 15 configurations I enumerated above are basically $6 \choose 4$ and one can see this if the consonants are placed first. See comment by @N.F. Taussig below.