Best Answer:
11!/(2!3!1!2!2!1!) = 831,600.

Take the number of ways to arrange the 11 digits if they were are different (11!) and divide out all of the ways that the the 1's, 2's, 4's, and 5's can be rearranged since all of those are the same numbers, and rearranging them yields the same number.

edit: I didn't see the even part. That means there will be less ways than this. OK, that means either

1. it ends with a two, which means that there are 1*10!/(2!2!1!2!2!1!) = 226,800 ways. This is because there is only one way for the last number to come one (as a 2) the other 10 digits can be arranged in 10!/(2!2!1!2!2!1!) ways. You multiply because of the General Counting Principle.

2. it ends with a four, which means that there are 1*10!/(2!3!1!1!2!1!) = 151,200 ways. Same reasoning as above.

3. it ends with a eight, which means that there are 1*10!/(2!3!1!2!2!) = 75,600 ways. Same reasoning as above.

Add these up. 453,600 ways.

edit: I just have to add that since 6/11 of the numbers that you are sorting are even, you should expect that 6/11 of the numbers that can be made are odd. Well

453,600/831,600 = 6/11.

I never really thought about that before. Pretty cool (at least to me).

edit: pi r squared, that would be correct if all 11 numbers are different. They are not and you have to take that into account.

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