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I have this problem in which a word (lets say "ABERRATIONAL") is given. Now how many permutations (no unique words) can be made given that the 3 A's have to be next to each other?

My solution: regard the 3 A's as one block, so we have 12-3+1 = 10 letters. This gives 10! permutations. The 3 A's yield 3! permutations. Answer = 10!.3!

This is wrong. The answer should be 302.400.

Help and insight is appreciated!

regards,

Marc

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# Permutation question

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