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 ?
- 124 - 1false
- 22425-1false
- 3(24-1)(23-1)25true
- 4Nonefalse
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Answer : 3. "(24-1)(23-1)25"
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 ballsor 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 ballsor 2 red balls from 3 red ballsor 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 ballsor 2 blue balls from 5 blue ballsor 3 blue balls from 5 blue ballsor 4 blue balls from 5 blue ballsor 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= 2524-123-1