Tough combinatorics problem

stop sine

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How many 5 letter passwords can be created using 3 A's and 2 B's? Order matters. Thanks for any help
 
Hello, stop sine!

How many 5-letter passwords can be created using 3 A's and 2 B's? Order matters.

There are 5 blanks to fill:   _  _  _  _  _\displaystyle \text{There are 5 blanks to fill: }\;\_\;\_\;\_\;\_\;\_

Choose 3 of them for the A’s. There are: (53)=5!3!2!=10 ways.\displaystyle \text{Choose 3 of them for the A's. }\:\text{There are: }\:{5\choose3} \:=\:\frac{5!}{3!2!} \:=\:10\text{ ways.}

Then drop the 2 B’s in the remaining spaces.\displaystyle \text{Then drop the 2 B's in the remaining spaces.}

. . Therefore, there are 10 possible passwords.\displaystyle \text{Therefore, there are }\boxed{10}\text{ possible passwords.}

 
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