Random variable and probability distribution: (5 men and 5 women are ranked) An exercise in my homework:
5 women and 5 men are ranked based on their results in a test. Assume that all results are different and every of 10! possible orders has the same probability. 
X (random variable) is the place of the highest ranked woman on the test. So it's possible values are 1,2,3,4,5,6.
Find the probability distribution.
So my approach is: 
$$\frac{(x-1)! \cdot 5 \cdot \left({(10-x)!}\over{((10-x)-(5-x+1))!}\right) \cdot 4!}{10!}$$
My thoughts:


*

*$10!$ : are the possible cases

*$(x-1)!$ :  are the possibilities to sort every man that is higher ranked than the first woman.

*5 : there are five woman so every woman can be the highest ranked.

*$\left({(10-x)!}\over{((10-x)-(5-x+1))!}\right) $ : all men behind the best woman can be placed on all the left places (not taken by the best woman or better men).

*$4!$ : are the possibilities to place the last 4 women on the left places.


To calculate each probability. Now my questions are:


*

*Am I right?

*Is there a simpler way to achieve this / can I simplify it?

The probability that $X=1$ is $1/2$ by symmetry.
The probability that the top ranked woman is second overall is the probability that the first is male and the second is female. This is $\frac{5}{10}\cdot\frac{5}{9}$.
The probability that the top ranked woman is third overall is the probability that the first two are male and the third is female. This is $\frac{5}{10}\cdot\frac{4}{9}\cdot \frac{5}{8}$.
Continue.
