A bag contains 5 black and 6 red balls.

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