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 $x_{3} < x_{4} < x_{5} ?$

• Correct answer
  A

  $1 / 120$
• B

  $1 / 60$
• C

  $1 / 20$
• D

  $1 / 10$

Solution

$\frac{100^{10} C_{5}[1 \times 2(3)]}{{ }^{100} \mathrm{C}_{5} \times 5 !}$ 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

Hence, the correct answer is option: 'A'.