There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is
A
6(7!−4!)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
7(6!−4!)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
8!−5!
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is D6(7!−4!)
These are totally 9 balls of which 2 are identical of one kind, 3 are a like of another kind and4 district ones.
At least one ball of same color separated = Total − No ball of same color is separated
Total permutation =9!2!3!
For no ball is separated : we consider all balls of same color as 1 entity, so there are 3 entities which can be placed in 3! ways.
The white and red balls are identical so they will be placed in 1 way whereas green balls are different so they can be placed in 4! ways