Problem relating to dividing persons in a group The question states as follows

There are only two women among 20 persons taking part in trip. The 20 persons are divided into 2 groups each consisting of 10 persons . Than the probability that two women will be in the same group is :

My method 


*

*Total ways = $(20!)/(10!×10!×2!)$ means by dividing them in 2 groups of 10 . 

*Favourable ways = $(18!)/(10!×8!)$ bydividing 18 persons in 2 groups of one 8 and other 10 persons . 


So as to send 2 women with group of 8 persons dividing these I get $9/19$ as the answer. But given answer is $9/38$ . Can someone tell what I have missed and what is the correct approach
 Thanks
 A: You are correct. 
Suppose everyone picks up a ticket giving their group from one of 20 face down on the table. One woman is the first to pick up a ticket. The chance that tne other picks up a matching ticket is $$\frac{9}{19}$$
A: Your solution is correct but somehow begs for a more simple approach.

Let's call the women Alice and Beatrice.
By the formation of groups Alice will be joined by $9$ of the other $19$ persons.
What is the probability that Beatrice is among these $9$ persons?
This already "smells" very much as $\frac9{19}$, but let us formalize a bit.
You can think of this as $9$ choices that are to be made out of $19$ persons. 
For every $i\in\{1,\dots,9\}$ all $19$ persons have equal chances to be chosen at the $i$-th choice.
So if $B_i$ denotes the event that Beatrice is chosen at the $i$-th choice then:$$\{\text{Beatrice will join Alice}\}=\bigcup_{i=1}^9B_i$$and consequently:$$P(\text{Beatrice will join Alice})=P\left(\bigcup_{i=1}^9B_i\right)=\sum_{i=1}^9P\left(B_i\right)=\sum_{i=1}^9\frac1{19}=\frac9{19}$$
