A bag contains 50 tickets numbered 1, 2, 3, . . ., 50 of which five are drawn at random and arranged in ascending order of magnitude 

$\left(x_1<x_2<x_3<x_4<x_5\right)\text{. }$. The probability that x₃ = 30 is

$$\begin{gathered} \text{ Five tickets out of }50{\text{ can drawn in }}^{50}{C}_5 \mathrm{ways} \\ \text{ Since }{x}_1<x_2<x_2<x_4<x_5\text{ and }{x}_2=30, \\ {x}_1,x_2<30\text{, i.e. }{x}_1\text{ and }{x}_2\text{ should come } \end{gathered}$$

$\text{ from tickets numbered }1\text{ and }29{\text{ and this may happen in }}^{29}{C}_2\text{ways}$

$\text{ Remaining ways, i.e. }{x}_4,x_5>30\text{, should come from }$

$20\text{ tickets numbered }31\text{ to }50{\text{ in }}^{20}{C}_2\text{ ways. }$

$\text{ So, favourable number of cases }{=}^{29}{{\mathrm{C}}_2}^{20}{\mathrm{C}}_2$

$\text{ Hence, required probability }=\frac{{ }^{29}{C}_2{\times }^{20}{C}_2}{{ }^{50}{C}_5}$