Combinatorics: Arrangement of letters with no isolated vowels

Question

I am studying combinatorics and I am doing a practice question without a solution key, I am not sure if I am doing correctly so plz help me check if my understanding is correct.

I will omit the full description of this problem, but the problem goes like this:

We have 8 letters, letters are formed by consonants and vowels, we have 3 specific consonants like {b, c, d}, and 5 vowels but only 4 types like {a, e, i, o}, but with two appearance of vowel e, so the vowels set look like {a, e, e, i, o}, the letter e is indistinguishable.

The question asks to compute the total number of possible arrangement of these 8 letters, but no isolated vowels, meaning any vowels in the arrangement must have another vowel beside it.

So my current understanding is this, I know vowel must be at least a pair to appear in the arrangement, so the total possible pair of vowels should have length like (2, 3), (3, 2), or just a chunk of length 5, and I know in the scenario when the chunk has length five, it should be something that looks like:

(v,v,v,v,v,c,c,c)

and I have managed to figure out that there are 4 possible locations for the chunk to sit in the arrangement, so I know in this case the number of total arrangements is $4* \frac{5!}{2!} * 3!$. (Please corrects me if I am doing this wrong).

My confusion arises for the (2, 3) and (3, 2) situations. I kind of have this feeling that the total possible arrangements are:

1. (v,v,c,c,c,v,v,v)
2. (c,v,v,c,c,v,v,v)
3. (c,c,v,v,c,v,v,v)
4. (v,v,c,c,v,v,v,c)
5. (v,v,c,v,v,v,c,c)
6. (v,v,v,c,c,c,v,v)
7. (c,v,v,v,c,c,v,v)
8. (c,c,v,v,v,c,v,v)
9. (v,v,v,c,c,v,v,c)

10.(v,v,v,c,v,v,c,c)

Here is my confusion, I am not sure if I am doing this correctly, but if I do, how do I use a numerical notation to indicate the above 10 scenarios? And why? Moreover, I feel like in every single of the conditions above, the result is all $\frac{5!}{2!} * 3!$, so the final solution should be $14 *\frac{5!}{2!} * 3!$? I feel like it's a wrong answer but I don't know where my mistake is, can anyone help me clarify this?

Thanks. :)

Answer 1

We can answer this with the intution of consecutive "choices". For the $5$-chunk case, there are indeed $4$ places for the chunk, so $4\frac{5!}{2!}3!$ possibilities.

Suppose we have a chunk of $3$ and a chunk of $2$. There must be at least one consonant between them. First, we choose which chunk to put first (2 possibilities). Next, we have 2 remaining consonants, and we can put them before, between, or after the chunks. By Bose-Einstein, there are ${2+3-1\choose 2}={4\choose 2}=6$ possibilities, meaning there should be $12$ arrangements. It looks like you missed $(c,v,v,c,v,v,v,c)$ and $(c,v,v,v,c,v,v,c)$. This gives a total of $(12+4)\frac{5!}{2!}3!$ words.