Question

How many permutations of the seven letters A, B, C, D, E, F, G do not have two consecutive vowels on the ends

Answer 1

The number of permutations without consecutive vowels at either end will be $4560$

Let $X = \{A, B, C, D, E, F, G\}$

Let $Perm(X)=\{\text{All possible permutations of the elements of X}\}$

Let $V = \{A, E\}$

Let $Perm(\overline{V})= \{\text{All possible permutations of the consonants in X}\}$

Permutations with consecutive vowels at either end will have one of the following forms

$AE\overline{v} \text{ or } EA\overline{v} \text{ or } \overline{v}AE \text{ or } \overline{v}EA\\\text{for } \overline{v} \in Perm(\overline{V})$

In all, we will have

$4\times |Perm(\overline{V})|=4\times 5!$  permutations

So, the number of permutations without consecutive vowels at either end will be

$|Perm(X)|-4\times|Perm(\overline{V})|=7!-4\times 5!=4560$

Another solved problem on permutations can be found here: https://brainly.com/question/25642091