There are 6 other positions in the string, each with 26 choices. So if you fix BO as the first two letters, there are [tex]26^6[/tex] possible strings that you can make.
If BO is at the end of the string, you still have [tex]26^6[/tex] possible strings.
Together, then, you have [tex]2\times26^6[/tex] possible strings.
