How many arrangements of the letters AABBCCDDEE are there such that there are no double letters in the sequence?
I am not positive, but would this not be...
10!/2!2!2!2!2!-(5*9!/2!2!2!2!-4*(8!/2!2!2!-3*(7!/2!2!-2*(6!/2!))))
= 68400 arrangements
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Posted by Dej Mar
on 2006-05-17 10:06:49 |