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. 
 A: 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 $\frac9{21}=\frac37$.
A: For the $i$-th ticket (when arranged in order) to have value $s$, you would need $i-1$ values less than $s$, and $r-i$ values greater than $s$ (of course you must also choose ticket $s$--one way to do that).  The numerator of your expression shows how many ways there are to do each of these, and then they are multiplied together in accordance with the multiplication rule.
