Tough combinatorics problem

stop sine

New member
Joined
Sep 9, 2008
Messages
1
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.

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

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

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

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

 
Top