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There are 2 white 3 black and 4 red balls. The no. of ways in which these balls can be arranged so that no two red balls can be together is? (Two balls of same colour are identical.)

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  • No. of arrangement of n things out of which p are of one category and q are of another category is $\large\frac{n!}{p!.q!}$
Since no conditions for white and black balls,
2 white and 3 black balls can be arranged in $\large\frac{5!}{2!.3!}$$=10 \:ways$
Since no two red can come together, red balls are to be placed
in between these 5 balls .(2 white,3 black)
There are 6 places and 4 balls as given below
$ - B_1 - B_2 - B_3 - B_4 -B_5 -$
Selection of any 4 places out of 6 places is $^6C_4=15\:ways$
$\therefore$ The required no. of arrangements $= 10\times 15=150$


answered Aug 31, 2013 by rvidyagovindarajan_1
edited Dec 23, 2013 by meenakshi.p

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