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Question

(B) The total number of ways to arrange $7$ white balls and $3$ black balls in a row is given by the combination formula $\binom{10}{3} = \frac{10!}{7

Vedclass · Accepted Answer

$\frac{7}{15}$

Vedclass · Answer

(B) The total number of ways to arrange $7$ white balls and $3$ black balls in a row is given by the combination formula $\binom{10}{3} = \frac{10!}{7!3!} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.To ensure no two black balls are adjacent,we first arrange the $7$ white balls in a row. This creates $8$ possible spaces (gaps) where the $3$ black balls can be placed (one before the first ball,one after the last ball,and $6$ between the balls).The number of ways to choose $3$ spaces out of $8$ is $\binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.Therefore,the required probability is $\frac{56}{120} = \frac{7}{15}$.

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

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