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:


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)


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. :)

  • $\begingroup$ Maybe try a smaller example where you can actually write out all the possibilities, and see if it matches your method here. $\endgroup$
    – coffeemath
    Sep 20, 2022 at 15:17
  • $\begingroup$ 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. $\endgroup$
    – MeTAZOIS
    Sep 20, 2022 at 15:20

1 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.

  • 1
    $\begingroup$ Thanks for clarifying the two options I didn't come up with, yeah I understand the final answer now, but I still feel confused about the combination you wrote after the Bose-Einstein, why is that? What are the twos things we select out of the 4? What is this 4? $\endgroup$
    – MeTAZOIS
    Sep 20, 2022 at 15:35
  • $\begingroup$ 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 $\endgroup$
    – Logan Post
    Sep 20, 2022 at 16:13

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