How many different permutations of this set don't have vowels on the ends? If we have the set of seven letters: (A,B,C,D,E,F,G) then how many permutations of these seven letters do not have vowels on the ends (that is, both the first and last letters are consonants)?  I was thinking 7!/(7!-2!) but I'm not sure how correct this is.  Thanks!
 A: You want to choose both end letters first: There are $5$ choices for the first end, and after that's chosen, you have $4$ remaining choices for the other end letter. The rest of the letters can be anything you want (and there are 5 choices to make), so the number of choices you have is $5\cdot 4\cdot 5! = 2400$.
Clarification: I'm interpreting "not having vowels on the ends" as meaning "both the first and last letters are not vowels."
A: Choose the end-letters: you can do so in $2\cdot\binom{5}{2}$ ways (select two of five non-vowels, then select in which end each one goes), then sort the rest as you want. $2\cdot\binom{5}{2}\cdot 5! = 2400$.
A: This is simple. 
How many permutations contains a specific symbol to a specific location (must be last) where the total number of characters (e.g. letters) is the n? After all (n-1)!  If this particular place can be one of two characters, then the number of permutations that are in a specific location, one or second character is $2x$ the same number, i.e.. $2(n-1)!$.
If count of letters is $n$ and count of vowels is $v$, then count of permutation ends with vowel is $v(n-1)!$ and, which is your question, count of permutations ends without vowels is $$(n-v)(n-1)!$$. Set of letters (A, B, C, D, E, F, G) contains 2 vowels (A, E). So your search count of permutations is $${5}\cdot{(7-1)! = 3600}$$.
