# How many permutations of this set can be made?

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
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.
@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