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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) 2425-1 C) (24-1)(23-1)25 D) None

Correct Answer: C

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
= 2 4 - 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
= 2 3 - 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
= 2 5 ..............(C)


 


From (A), (B) and (C), required number of ways
=   2 5 2 4 - 1 2 3 - 1


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