# Combinatorics: Arrangement of letters with no isolated vowels

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. :)

• Maybe try a smaller example where you can actually write out all the possibilities, and see if it matches your method here. Sep 20, 2022 at 15:17
• I am aware of examples with smaller samples that I can list all the possibilities, it's just this complex one I am just having no clue how to use numerical notation to denote each case. Sep 20, 2022 at 15:20

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.
• Bose-Einstein counts the number of ways to distribute n indistinguishable objects into k distinguishable spots. In this case, counting the ways to place n=2 consonants into k=3 spots. The formula is ${n+k-1\choose n}$. Look here fore explanation link Sep 20, 2022 at 16:13