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Arithmetic Ability
Permutations and Combinations

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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) 24 − 1 B) 24(25−1) C) (24−1)(23−1)25 D) None

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Correct choice: 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 = 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)
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