seven white balls and three black balls are randomly placed in a row . the probability that no two black balls are placed adjacently equal Share with your friends Share 1 Shruti Tyagi answered this Dear student, Placing 7 white balls in a row leaves 8 gaps. The 3 black balls can be placed in 8 gaps =8×7×63×2×1=56ways So, the total number of ways of arranging white and black balls such that no two black balls are adjacent = 56×3!×7! Actual number of arrangements possible with 7 white and 3 black balls = (7+3)! = 10! Therefore probability that no two black balls are adjacent =56×3!×7! 10!=75×3=715 Regards 2 View Full Answer Himanshu Sharma answered this Here we can use permutation ... First fix the seven white balls, then you have eight places for the black balls to choose from... and 8C3 is the number of combinations you can have ( how the places are chosen) Total number of arrangements of the 10 balls is 10C3 So prob = 8C3 / 10C3 =?7/15 2 Pulkit Goyal answered this yo 0
