Answer
$P(selecting~two~60-watt~bulbs)=\frac{153}{435}\approx0.3517$
Work Step by Step
Combinations of 30 distinct light bulbs (40-watt and 60-watt) taken 2 at a time (the order in which the light bulbs are selected does not matter):
$N(S)=~_{30}C_2=\frac{30!}{2!(30-2)!}=\frac{30!}{2!\times28!}$
But, $30!=30\times29\times(28\times27\times26\times...\times3\times2\times1)=30\times29\times28!$
$_{30}C_2=\frac{30\times29\times28!}{2!\times28!}=\frac{30\times29}{2}=435$
There are 18 60-watt light bulbs.
Combinations of 18 distinct 60-watt light bulbs taken 2 at a time (the order in which the 60-watt light bulbs are selected does not matter):
$N(selecting~two~60-watt~bulbs)=~_{18}C_2=\frac{18!}{2!(18-2)!}=\frac{18!}{2!\times16!}$
But, $18!=18\times17\times(16\times15\times14\times...\times3\times2\times1)=18\times17\times16!$
$_{18}C_2=\frac{18\times17\times16!}{2!\times16!}=\frac{18\times17}{2}=153$
Using the Classical Method (page 259):
$P(selecting~two~60-watt~bulbs)=\frac{N(selecting~two~60-watt~bulbs)}{N(S)}=\frac{153}{435}\approx0.3517$
