Question

A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball ?

A) 24 - 1 B) 24(25-1) C) (24-1)(23-1)25 D) None

Answer 1

Explanation:

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

 

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

 

Hence, we can select 1 black ball from 4 black balls

or 2 black balls from 4 black balls.

or 3 black balls from 4 black balls.

or 4 black balls from 4 black balls.

 

Hence, number of ways in which we can select the black balls

 

= 4C1 + 4C2 + 4C3 + 4C4

= 24−1 ........(A)

 

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

 

Hence, we can select 1 red ball from 3 red balls

or 2 red balls from 3 red balls

or 3 red balls from 3 red balls

 

Hence, number of ways in which we can select the red balls

= 3C1 + 3C2 + 3C3

=23−1........(B)

 

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)

or 1 blue ball from 5 blue balls

or 2 blue balls from 5 blue balls

or 3 blue balls from 5 blue balls

or 4 blue balls from 5 blue balls

or 5 blue balls from 5 blue balls.

 

Hence, number of ways in which we can select the blue balls

= 5C0 + 5C1 + 5C2 + … + 5C5

= 25..............(C)

 

From (A), (B) and (C), required number of ways

=  25(24−1)(23−1)

ANSWER IS C...