Urn containg tickets problem From an urn containing n tickets numbered $1, 2, \dots, n$, $r$ tickets are drawn simultaneously and arranged in increasing order of their numbers $x_1<x_2<\dots<x_r$. Show that the probability that $x_i=s$ is
$$\frac{{s-1 \choose i-1}{n-s \choose r-i}}{{n \choose r}}$$
My attempt
From n tickets, r tickets can be drawn in ${n \choose r}$ ways. From the question, I understand that after drawing the tickets they should be arrangers in increasing order and so their  positions cannot be interchanged. Now I am unable to understand the meaning probability that $x_i=s$.
Please explain the meaning of the question and help me to solve. 
I hope someone can help. Thanks. 

There $\binom{n}{r}$ possibilities when $r$ tickets are drawn without any further conditions. The condition $x_i=s$ can be translated into: from the tickets having a number smaller than $s$ exactly $i-1$ must be drawn and from the tickets  having a number greater than $s$ exactly $r-i$ must be drawn (then automatically ticket with number $s$ will be drawn and this with $x_i=s$). This gives $\binom{s-1}{i-1}\binom{n-s}{r-i}$ possibilities.

addendum (example):
Suppose $n=7$, $r=5$, $s=4$ and $i=3$. Then we will get $x_3=4$ in the following cases:


*

*$1,2,4,5,6$ (note that here $x_1=1, x_2=2, x_3=4, x_4=5$ and $x_5=6$)

*$1,3,4,5,6$

*$2,3,4,5,6$

*$1,2,4,5,7$

*$1,3,4,5,7$

*$2,3,4,5,7$

*$1,2,4,6,7$

*$1,3,4,6,7$

*$2,3,4,6,7$


The first $2$ comes to choosing $2$ out of $\{1,2,3\}$, so $\binom{s-1}{i-1}=\binom32=3$ possibilities.
The last $2$ comes to choosing $2$ out of $\{5,6,7\}$, so $\binom{n-s}{r-i}=\binom32=3$ possibilities.
That gives $3\times3=9$ possibilities. 
There are $\binom{n}{r}=\binom75=21$ possibilities in total, so the probability that one of the mentioned possibilities will occur is $\frac9{21}=\frac37$.
