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There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number of possible ways in which they can be arranged in a row so that atleast one ball is seperated from the balls of the same colour is ?

There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number of possible ways in which they can be arranged in a row so that atleast one ball is seperated from the balls of the same colour is ?

Grade:11

2 Answers

Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
9 years ago
Hi Student

The above can be solved as total cases minus the number of cases in which no ball is separated

Total cases- 9!/2!3! ( as green ball are of different shades they can be distinguished)
Number of cases in which all are together is by considering 2 while ball as one unit 3 red balls as one unit and 4 green balls as one unit is 3!

Required ways is 9!/2!3! – 3!
KUNAL PATEL
57 Points
6 years ago
SIR(HARSH PATODIA)
YOUR ANSWER IS ALMOST CORRECT BUT YOU JUST MADE A MISTAKE. WHEN YOU CONSIDERED ALL COLOURS BALLS AS ONE UNIT THEN POSSSIBLE PERMUTATION WILL BE 3!.4! (4! IS BECAUSE GREEN BALLS ARE OF DIFFERENT SHADES SO THERE SHOULD BE INTRA PERMUTATION.SO THE FINAL ANSWER IS 
9!/2!3! – 3!.4!=30096

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