Question : A bag contains 40 tickets numbered $1, 2, 3, ...,40$ of which four are drawn at random and arranged order $(t_1 < t_2 < t_3 < t_4)$. Find the probabitlity of $t_3$ being $20$.
My solution was: If $t_3 = 20$, then the tickets $t_1$ and $t_2$ must come out from $19$ tickets(1 to 19). This can be done in $C(19, 2)$ ways. $t_4$ must come from $20$ tickets ($21$ to $40$), which can be done in $C(20, 1)$ ways. So Ans $ = C(19, 2) \times C(20, 1)$
But I checked the solution in book, it states that total number of cases$ = C(40, 4)$ ways I Agree
Therefore favorable number of cases(i.e. solution) is $(C(19, 2) \times C(20,1)) / (C(40, 4))$
I want to know why extra division by whole possible cases $C(40, 4)$ is done to get the solution of problem. And what number does my solution represents