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(C) The total number of ways to select and arrange $5$ coupons from $100$ is $P(100, 5) = \frac{100!}{95!}$.Let the set of $5$ chosen numbers be $S =

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$1 / 20$

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(C) The total number of ways to select and arrange $5$ coupons from $100$ is $P(100, 5) = \frac{100!}{95!}$.Let the set of $5$ chosen numbers be $S = \{a, b, c, d, e\}$ where $a For any set of $5$ distinct numbers,we need to arrange them such that $x_1 > x_2 > x_3$ and $x_3 This implies $x_3$ must be the smallest of the $5$ numbers. Thus,$x_3$ is fixed as the minimum element of the set.The remaining $4$ numbers can be partitioned into two sets: one for ${x_1, x_2}$ and one for ${x_4, x_5}$.The number of ways to choose $2$ numbers out of the remaining $4$ to be $x_1$ and $x_2$ is $\binom{4}{2} = 6$.Once chosen,their order is fixed $(x_1 > x_2)$,and the order of the remaining $2$ is also fixed $(x_4 Thus,for any set of $5$ numbers,there are $6$ favorable arrangements out of $5! = 120$ total permutations.The probability is $\frac{6}{120} = \frac{1}{20}$.

$A$ box contains coupons labelled $1, 2, \ldots, 100$. Five coupons are picked at random one after another without replacement. Let the numbers on the coupons be $x_1, x_2, \ldots, x_5$. What is the probability that $x_1 > x_2 > x_3$ and $x_3 < x_4 < x_5$?

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