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?

Answer 1

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  ⋯=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  ⋯=2^3−1  ⋯(B)

Now let's find out the number of ways in which we can select the blue balls. There is no specific condition given here.

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  ⋯=2^5  ⋯(C)

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