Calculating pm problem: $5$ man and $5$ women running in a race Let's say there are $5$ man and $5$ women running in a race. Let the variable X that indicates the highest position in ranking for a woman. So, for example, if $X = 1$, that means that a woman took first place(any woman) and if $X = 6$, that means that women took the last places on the ranking. 
I guess, obviously there are $10!$ different outcomes for ranking. 
How can I calculate $p_X(x)$ with $x = 1, 2, 3, 4, 5, 6$?
Edit : Two or more people cannot take the same place in the ranking. All 10! different outcomes are equally possible. 
 A: The women are assumed to be indistinguishable. There are ${10 \choose 5}$ ways to place them in $10$ empty slots. You're correct that $6$ is highest value $X$ can take on. 
If we are assuming every one of the $10$ people are equally fast, then the probability that a woman is in the first position is $\frac{1}{2}$ by symmetry.
The probability that the best position by a woman is $X=2$ is $$\frac{5}{10}\cdot\frac{5}{9}=\frac{{5 \choose 1}}{{10 \choose 1}}\cdot\frac{{5 \choose 1}}{{9 \choose 1}}$$
since a man has to be in the first position and then a woman has to be in the second position
The probability that the best position by a woman is $X=3$ is $$\frac{5}{10}\cdot\frac{4}{9}\cdot\frac{5}{8}=\frac{{5 \choose 2}}{{10 \choose 2}}\cdot\frac{{5 \choose 1}}{{8 \choose 1}}$$
since a man has to be in the first and second position and then a woman has to be in the third position
$\vdots$
The probability that the best position by a woman is $X=6$ is $$\frac{5}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{1}{6}\cdot\frac{5}{5}=\frac{{5 \choose 5}}{{10 \choose 5}}\cdot\frac{{5 \choose 1}}{{5 \choose 1}}$$
Can you go from here to obtain $p_X(x)$?
A: Here is a solution where order matters.
$P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)$
$=\frac{{5 \choose 1}\times9!}{10!}+\frac{{5 \choose 1}\times{5 \choose 1}\times8!}{10!}+\frac{{5 \choose 2}\times 2! \times{5 \choose 1}\times7!}{10!}+\frac{{5 \choose 3}\times 3! \times{5 \choose 1}\times6!}{10!}+\frac{{5 \choose 4}\times 4! \times{5 \choose 1}\times5!}{10!}\frac{{5 \choose 5}\times 5! \times{5 \choose 1}\times4!}{10!}$
$=\frac{5}{10}+\frac{5\times5}{10\times9}+\frac{5\times4\times5}{10\times9\times8}+\frac{5\times4\times3\times5}{10\times9\times8\times7}+\frac{5\times4\times3\times2\times5}{10\times9\times8\times7\times6}+\frac{5\times4\times3\times2\times1\times5}{10\times9\times8\times7\times6\times5}$
