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I was given this question:

Find the number of ways all $10$ letters of the word COPENHAGEN can be arranged so that the Es are not next to each other.

This is how I tried to solve it:

  • Permutations with only one E $= 9!/2! = 181~440$

  • 2nd E can be in any of $8$ places

(2nd E can be any where with # [when L stands for a letter that is not E] #L#L#L#L E L#L#L#L# this counts for any arrangement for instance if E is on the edge E L#L#L#L#L#L#L#L#)

  • Then total number of ways $= 9!/2! \cdot 8 = 1~515~520$

However the correct answer is $725~760$... which is half of my answer....

I am aware of the other method for solving this (using [total ways to arrange all $10$ letters] - [ways to arrange letters when Es are together) but I want to know where I went wrong with this method.

Can anyone explain why I have to divide my answer by $2$ for it to be right?

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    find the number of arrangements of COPENHAGEN and subtract the number arrangements of COPENHAGN. – 2017-02-23

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You consider the E's to be distinct, which results in your answer being twice the correct one. I can infer that because you used the phrase "2nd E".