Question 467690
**For any set of n objects, there are n! ways of arranging them**
Now for the letters A-F, there are 2 vowels and 4 consonants 
If a vowel is at one end and a consonant at the other then that leaves 4 letters remaining to be arranged in any which way between them.
From above we know then that are 4! ways of arranging 4 letters.
Say the consonant B is on the front end and a vowel on other end:
B _ _ _ _ Vowel
There are only 2 possibilities because there are only 2 vowels to choose from:
B _ _ _ _ A     OR   B _ _ _ _ E
Each of these has 4! arrangements
So with B at the front there are 2*(4!) arrangements
However, there are 4 consonants each of which could be at the front end
So multiply by 4
With any consonant at front end there are 4*2*(4!) arrangements
However, the consonant could also be at back end with vowel in front
Since everything else would stay the same, double the possible arrangements
Total number of arrangements are 2*4*2*(4!) arrangements which simplifies to 16(4!)