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 from the lot.
Solution:
We know that,
nCr
$=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}$
According to the question,
Number of black balls = 5
Number of red balls = 6
Number of ways in which 2 black balls can be selected = 5C2
$=\frac{5 !}{2 ! 3 !}=10$
Number of ways in which 3 red balls can be selected =5C3
$=\frac{6 !}{3 ! 3 !}=20$
Number of ways in which 2 black & 3 red ball can be selected
=5C2×5C3
=10×20
=200