A bag contains 50 tickets numbered 1,2,3……, 50 of which five are drawn at random and arranged in ascending order of magnitude $\left({\mathrm{x}}_1<{\mathrm{x}}_2<{\mathrm{x}}_3<{\mathrm{x}}_4<{\mathrm{x}}_5\right)$. The probability that $x_3=30$   is

Since, ${\mathrm{x}}_1<{\mathrm{x}}_2<{\mathrm{x}}_3<{\mathrm{x}}_4<{\mathrm{x}}_5\text{ and }{\mathrm{x}}_3=30$, Therefore, ${\mathrm{x}}_1,{\mathrm{x}}_2<30\text{, i.e. }{\mathrm{x}}_1\text{ and }{\mathrm{x}}_2$ should come from tickets numbered 1 to 29 and this may happen in ${ }^{29}{\mathrm{C}}_2$ ways.
${\mathrm{x}}_4,{\mathrm{x}}_5>30$, should come from 20 tickets numbered 31 to 50 in ${ }^{29}{\mathrm{C}}_2$ ways.
$\therefore \mathrm{n}\left(\mathrm{A}\right){=}^{29}{\mathrm{C}}_2{\times }^{20}{\mathrm{C}}_2,\mathrm{n}\left(\mathrm{S}\right){=}^{50}{\mathrm{C}}_5\Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\frac{{ }^{29}{\mathrm{C}}_2{\times }^{20}{\mathrm{C}}_2}{{ }^{50}{\mathrm{C}}_5}$