Permutation and CombinationHard
Question
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 :
Options
A.6 (7 ! - 4!)
B.7 (6 ! - 4 !)
C.8! - 5!
D.none
Solution
Total number of ways of arranging 2 identical white balls.
3 identical red balls and 4 green balls of different shades =
= 6.7!
Number of ways when balls of same colour are together = 3! × 4! = 6.4!
∴ Number of ways of arranging the balls when atleast one ball is separated from the balls of the same colour = 6.7! - 6.4! = 6(7! - 4!)
3 identical red balls and 4 green balls of different shades =
Number of ways when balls of same colour are together = 3! × 4! = 6.4!
∴ Number of ways of arranging the balls when atleast one ball is separated from the balls of the same colour = 6.7! - 6.4! = 6(7! - 4!)
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