## Algebra 2 (1st Edition)

a) There are $9!$ different permutations of the musicians, and out of these only $1$ is in alphabetical order by last name. Thus the probability: $\frac{1}{9!}=\frac{1}{362880}\approx0.00000275573$ b) There are $_9C_2$ combinations of $2$ musicians and out of these only $_4C_2$ are your friends. Thus the probability: $\frac{_4C_2}{_9C_2}=\frac{6}{36}=\frac{1}{6}\approx0.16667$