Question

In hiw many ways can two ball of same colour be selected out from 4 black and three white balls

Answer 1

Answer:  The required number of ways is 18.

Step-by-step explanation:  We are given to find the number of ways in which two balls of same color can be selected out of 4 black and 3 white balls.

The number of ways in which 2 black balls are selected out of 4 black balls is given by

$n_1=^4C_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4\times3\times2!}{2\times1\times2!}=6$

and the number of ways in which 2 white balls can be selected out of 3 white balls is given by

$n_2=^3C_2=\dfrac{3!}{2!(3-2)!}=\dfrac{3\times2!}{2!\times1}=3.$

Therefore, the required number of ways in which two balls of same color can be selected is

$n=n_1\times n_2=6\times3=18.$

Thus, the required number of ways is 18.