Question:
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Solution:
There are 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in ${ }^{5} \mathrm{C}_{2}$ ways and 3 red balls can be selected out of 6 red balls in ${ }^{6} \mathrm{C}_{3}$ ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls
$={ }^{5} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{3}=\frac{5 !}{2 ! 3 !} \times \frac{6 !}{3 ! 3 !}=\frac{5 \times 4}{2} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}=10 \times 20=200$