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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!

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This question might win the medal for question with the most distinct answers. – Jonathan Rich Mar 20 '13 at 18:23
Lol yeah, very confusing.. – mario1433 Mar 20 '13 at 18:25
I think the difference is due to the requirement of "no vowel at the ends." By ends, do you mean the last letter of each permutation cannot be a vowel, vs. no vowel at each end: first and last letter of a permutation cannot be a vowel. – amWhy Mar 20 '13 at 18:29
Both ends of the set can contain no vowels* sorry! – mario1433 Mar 20 '13 at 18:31
up vote 3 down vote accepted

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."

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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. $2\cdot\binom{5}{2}\cdot 5! = 2400$.

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I believe he wants both the first and last letters not to be vowels. – Ian Coley Mar 20 '13 at 18:23
I just want there to be no vowels on the ends – mario1433 Mar 20 '13 at 18:28
@FrankMcGovern Thank you, I misread. Corrected. – Alfonso Fernandez Mar 20 '13 at 18:34

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 $${5}\cdot{(7-1)! = 3600}$$.

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