# How many even 11-digit numbers can be formed from the digits 1, 1, 2, 2, 2, 3, 4, 4, 5, 5, or 8?

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• 1 decade ago

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.

• 1 decade ago

Hi,

If you had 11 different things, they could be arranged in 11! different ways. ( 11! is eleven factorial which means multiply all the integers from 1 through 11 together to get the answer.)

But you want them to be even which means the last position always has to be an even number, any of the 6 even numbers, while the other 10 spots can be any of the remaining 10 digits. That would be any of 6 even choices for the last spot, then any of 10, any of 9, ... This would be 6 * 10! or 21,772,800 even numbers.

If you want them, it will take a long time to write them all!

I hope this helps.

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