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How many permutations of the set of seven letters (A,B,C,D,E,F,G) have the two vowels before the five consonants?

I'm wondering here if we use the set of 7! - 2! since they can only occupy the first two spaces?

In addition, I was also curious how many permutations have A immediately to the left of E?

If we consider that there are 7 spaces, minus 1 for the space next to E, that's 6!..but not sure if this is right. Thanks!

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How many ways are there to permute {A,E}? How many ways are there to permute {B,C,D,F,G}? How can I take a permutation of the first group and one of the second to make a permutation satisfying your constraint? –  A Blumenthal Mar 20 '13 at 18:41
    
{A,E} = 2!. To Permute {B,C,D,F,G} = 5!. To take permutation of the first group and one of the second is to..help me here haha –  mario1433 Mar 20 '13 at 18:43

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You know that in order to get the two vowels before the five consonants, you must put the vowels in the first two spaces, so that you have a string of the form VVCCCCC. You can build such a string in two parts: first assign the vowels $A$ and $E$ to the VV slots, and then assign the consonants $B,C,D,F$, and $G$ to the CCCCC slots. You know that there are $2!$ ways to perform the first part of the task and $5!$ ways to perform the second part. These parts are independent: no matter whether you place the vowels in the order $AE$ or in the order $EA$, you can place the consonants in any of the $5!$ possible orders. So how should you combine the numbers $2!$ and $5!$ to get the total number of possible strings?

Your answer of $6!$ to the second question, however, is right, as is your reasoning: you can treat the pair $AE$ as a single entity, so that you really have only $6$ things to arrange, and there are $6!$ ways of arranging them.

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Wait, if we treat AE as a single entity, aren't there 5 spaces left to move? which would mean 5! right? –  mario1433 Mar 20 '13 at 18:47
    
And do we add the 2! + 5!? Thanks! –  mario1433 Mar 20 '13 at 18:48
    
@mario1433: (1) No, you’re permuting $6$ things: $B,C,D,F,G$, and $AE$. Those $6$ things can go in any order. (2) No. Each of the $2$ arrangements of the vowels can be paired with any of the $120$ arrangements of the consonants, so you get $2\cdot120=240$ possible orders. This is the multiplication (or Chinese menu) principle, and it’s one of the most important tools you have for counting things. –  Brian M. Scott Mar 20 '13 at 18:56

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