Asked by Rinkus | 13th Sep, 2008, 01:00: PM
first we will arrange 6 white balls. total no. of ways of doing this = 6!
now we have 7 locations where we can place black balls ( five between white balls and one on left of leftmost white ball and one on right of rightmost white ball) so that no two black balls are together.
we have to choose 5 of them which is 7C5 and these black balls can be arranged by 5! ways .
so total no, of ways of doing this = 7C5(6!)(5!)
Answered by | 13th Sep, 2008, 05:28: PM
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