Q:
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)
B)
C)
D) None
Answer & Explanation
Answer: C) 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
= ........(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
=........(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
= ..............(C)
From (A), (B) and (C), required number of ways
=
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