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(B) Total number of ways to arrange $7$ white balls and $3$ black balls in a row is $\frac{10!}{7!3!} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes

Vedclass · Accepted Answer

$\frac{7}{15}$

Vedclass · Answer

(B) Total number of ways to arrange $7$ white balls and $3$ black balls in a row is $\frac{10!}{7!3!} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. To ensure no two black balls are adjacent,we place the $7$ white balls first,creating $8$ possible gaps (including ends) where the $3$ black balls can be placed. The number of ways to choose $3$ gaps out of $8$ is $\binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$. The required probability is $\frac{\binom{8}{3}}{\binom{10}{3}} = \frac{56}{120} = \frac{7}{15}$.

Seven white balls and three black balls are randomly arranged in a row. The probability that no two black balls are placed adjacently is

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