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I have a counting problem:

How many ways can the alphabet be rearranged so that A comes before B in a random order?

I know there are 26! ways for the alphabet to be arranged, but I'm not sure how to figure out the A before B part.

Any help is appreciated. Thanks!

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In half of the 26! permutations $A$ will be before $B$. So it is 26!/2.

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    So, I'm guessing that using the same logic, its 26!/2 for A before Z since it's a random order? Also, is the number of ways for A to come just before B 25!?2012-12-13
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    First question yes, for every ...A...Z... permutation there is one ...Z...A... 2nd Question Yes: Here AB can be seen as one single "supercharacter".2012-12-13
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    Thank you. I appreciate the help.2012-12-13