Number of permutations that strictly contain two consecutive vowels

Question

Given the string "ENGINEER", how might one find the number of permutations that contain only two consecutive vowels?

I understand that the total number of permutations is $\frac{8!}{3!2!}$, but to be honest, I'm not sure how to proceed with this problem.

Answer 1

Consider the vowel position mask - this needs to occupy 3 distinct slots in and around the four consonants. The consonants define a frame like:

$$\_c\_c\_c\_c\_$$

so we need to choose 3 of those 5 slots, ${5 \choose 3}$, then chose one of those chosen slots to be a double vowel, ${3 \choose 1}$. Then we can permute the four vowels and four consonants into the mask, taking account of duplicates: $\frac{4!}{3!}$ and $\frac{4!}{2!}$ respectively. This gives the final combinations meeting the criteria as:

$${5 \choose 3} {3 \choose 1}\frac{4!}{3!}\frac{4!}{2!} = 10\cdot 3\cdot 4 \cdot 12 = 1440$$

Answer 2

Hint: With eight slots and four vowels there are few patterns after you select the place the vowel pair is in. I would suggest you consider all the vowels and all the consonants distinct, solve the problem, and divide by the permutations of the identical letters, which you know to be $3!2!$