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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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Verified Answer
The correct answer is:
$1 / 120$
$\frac{100^{10} C_{5}[1 \times 2(3)]}{{ }^{100} \mathrm{C}_{5} \times 5 !}$
Suppose $1,2,3,4,5$ are
$$
=\frac{1}{20}
$$
Place of 1 is fixed
Total arrangements of 5 is 2
and
\begin{array}{|l|l|l|l|}
\hline & & 1 & & 5 \\
\hline
\end{array}
arrangements of $2,3,4$, are
$\left.\begin{array}{lllll}2 & 3 & 1 & 4 & 5 \\ 4 & 2 & 1 & 3 & 5 \\ 4 & 3 & 1 & 2 & 5\end{array}\right\}$ 3way
Suppose $1,2,3,4,5$ are
$$
=\frac{1}{20}
$$
Place of 1 is fixed
Total arrangements of 5 is 2
and
\begin{array}{|l|l|l|l|}
\hline & & 1 & & 5 \\
\hline
\end{array}
arrangements of $2,3,4$, are
$\left.\begin{array}{lllll}2 & 3 & 1 & 4 & 5 \\ 4 & 2 & 1 & 3 & 5 \\ 4 & 3 & 1 & 2 & 5\end{array}\right\}$ 3way
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