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The solution says that half of the total number of arrangements would have A before E . How is that?

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    can u provide your solution and source? whether the question is to find number of arrangements where A must come before E? what is the relevance of alphabetical order? – 2017-01-19
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    Well, half will have A before E and half will have E before A. What other possibilities would there be, and how would those two options have different numbers? – 2017-01-19

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Because the other half has E before A; you can pair them up two-by-two ("garden" with "gerdan", "ragdne" with "regdna", and so on) to show that there are equally many of each kind.

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    With 360 arrangements I'm finding it difficult to get the whole idea behind it. Could you give an example that narrows down the total number of arrangements? – 2017-01-19
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    @Sidd Try it out with "GAE" or "GARE". The are fewer combinations there. – 2017-01-19
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    For every xAyEzw There is another word xEyAzw which is *exactly* the same except the the E and the A are switcheed. No need to list them all. For gArdEn there is gErdAn. For drAgEn there is drEgAn... any time you have a ~~A~E~ you will have exactly one other ~~E~A~. That should be obvious, shouldn't it. – 2017-01-19
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    Or as arthur suggests, just use A,E,G,R for AEgr there is also EAgr. For AgEr there is EgAr. For AgrE there is EgrA. gAEr there is gEAr. For gArE ther is gErA. etc. – 2017-01-19
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Example.

Let you have two alphabets A, E to arrange you have two cases AE or EA. And you wants cases with A comes before E. Then you have only one way AE. That is half of the total cases.

Similar is with word "GARDEN".