# Ways to arrange a word so that no vowel is isolated between two consonants

Consider a seven-letter word formed by mixing up the letters in the word COMBINE. How many ways can you do this if no vowel is isolated between two consonants?
(eg. EBMCION and MOIENCB are acceptable, but BEMCNIO is not)

The final answer is given to be $1872$.

We are being asked to find the numeric value of (Total number of seven-letter words) - (Number of words such that a vowel is isolated between two consonants). The word "COMBINE" has $4$ consonants and $3$ vowels. To compute the number of constructed words with an isolated vowel between two consonants, I tried the following:

## 1st attempt:

We may generalize the case, representing consonants as $1$s and vowels as $0$s. Now, we are finding the total number of seven-digit binary strings with four $1$s and three $0$s, containing the substring $101$. There are $\frac{4!}{2!2!} = 6$ strings that can be formed from the remaining two $0$s and $1$s. We may insert the substring $101$ into any position in these 6 strings. We have five available positions, because there are four digits. Therefore, there are $6 \times 5=30$ seven-digit binary strings with four $1$s and three $0$s, containing the substring $101$.

Since each vowel and consonant is unique in the word COMBINE, there are $30 \times 4!\times3!=4320$ seven-letter words formed from COMBINE that contain an isolated vowel between two consonants, if we account for permutations of the vowels and consonants. However, $7!-4320=720 \neq 1872$, meaning my answer is incorrect.

## 2nd attempt:

Since there are three vowels, for there to be an isolated vowel between two consonants, either all vowels are separated, or two vowels may be grouped up together while the remaining vowel is isolated (there is no other case).

Case when all vowels are isolated:
First, we arrange the consonants, Then, we choose three out of five possible places to place the vowels (.C.C.C.C., where C denotes a consonant and the periods represent possible places to put a vowel). We finally account for the arrangement of the vowels. $$P(4,4) \times C(5,3) \times P(3,3) = 1440.$$

Case when one vowel is isolated, while the remaining two are a pair:
First, we arrange the consonants. We then choose a vowel. Then, we choose a position out of three possible places to put it (C.C.C.C). Then, we choose a position out of four possible places to place the pair of the remaining vowels (.CVC.C.C. just as a possible example). Finally, we account for the arrangement of the vowels in the pair. $$P(4,4) \times C(3,1) \times C(3,1) \times C(4,1) \times P(2,2) = 1728.$$ $7!-(1440+1728) = 1872$, which confirms that this is the correct answer.

What is wrong with my first attempt? Where did I make my error?

• For your first attempt, you are overcounting the bad strings. You count strings containing more than $1$ copy of $101$ multiple times.
– lulu
Commented Aug 7, 2016 at 12:27
The problem with the first attempt is that bad strings with multiple $101's$ are counted multiply. Thus $1010101$ is counted three times (once for the $101$ starting in the first slot, then again for the $101$ starting in the third, and then a third time for the $101$ starting in slot five). Not easy to repair.