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4 A's and 5 B's are to be arranged into a nine letter word. How many different words can be made?

 

I thought it would be (9!/something) but I don't know how to get the denominator.

Guest Mar 9, 2017
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We have :

 

9!  / [ (number of repeats of letter A) ! * (number of repeats of letter B) ! ]  =

 

9! / ( 5! * 4!)  =  126  identifiable words

 

 

cool cool cool

CPhill  Mar 9, 2017

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