A child arranges 3 distinct black balls, four green balls, and five red balls in a row randomly. If $p$ is the probability that two red balls are not together, find $99 p$ .

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Solution

total no ways of arranging $3$ black $4$ green $5$ red is $\frac{12!}{3! .4! .5!}$ .
given that no $2$ reds are adjacent so first arrange remaining such that leave place in between so that reds can be arranged in middle $\frac{7!}{3! .4!}$ .
now there are $8$ places available for $5$ reds we can select it in $8_{C_{5}}$ .
$p = \frac{12! \frac{12!}{3! .4! .5!}}{\frac{7!}{3! .4!} \times 8_{C_{5}}}$
$p = 7 / 99$

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A child arranges 3 distinct black balls, four green balls, and five red balls in a row randomly. If p is the probability that two red balls are not together, find 99p.

A child arranges 3 distinct black balls, four green balls, and five red balls in a row randomly. If $p$ is the probability that two red balls are not together, find $99 p$ .