Answer
$P(all~3~contain~diet~soda)=\frac{1}{220}\approx0.004545$
Work Step by Step
Combinations of 12 distinct cans (regular or diet soda) taken 3 at a time (the order in which the cans are selected does not matter):
$N(S)=~_{12}C_3=\frac{12!}{3!(12-3)!}=\frac{12!}{3!\times9!}=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{3\times2\times1\times9\times8\times7\times6\times5\times4\times3\times2\times1}=220$
Combinations of 3 distinct diet soda taken 3 at a time (the order in which the cans are selected does not matter):
$N(3~diet~soda~among~the~3~diet~soda)=~_{3}C_3=\frac{3!}{3!(3-3)!}=\frac{3!}{3!\times0!}=1$ (Remember: $0!=1$)
Using the Classical Method (page 259):
$P(all~3~contain~diet~soda)=\frac{N(3~diet~soda~among~the~3~diet~soda)}{N(S)}=\frac{1}{220}\approx0.004545$
