Correct logic of permuting 5 men and 5 women to find probability of different highest women rank The problem reads like this:

Problem
  Five men and $5$ women are ranked according to their scores on an examination. Assume that no two scores are alike and all $10!$ possible rankings are equally likely. Let $X$ denote the highest ranking achieved by a woman. (For instance, $X = 1$ if the top-ranked person is female.) Find $P(X = i),i = 1, 2, 3, . . ., 8, 9, 10$.
Solution given was:
  
  
*
  
*$P(X=1)= \frac{5}{10}= \frac{1}{2}$ because there are 5 women and total of 10 to choose from
  
*$P(X=2)=\frac{5}{10}\times \frac{5}{9}=\frac{5}{18}$ because for rank1 thereare 5 men and total of 10 to choose from, for rank 2 (we want awoman) we still have 5 women but only a total of 9 to choosefrom.
  
*$P(X=3)=\frac{5}{10}\times \frac{4}{9}\times \frac{5}{8}=\frac{5}{36}$
  
*$P(X=4)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{5}{7}=\frac{10}{168}$ 
  
*$P(X=5)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{5}{6}=\frac{5}{252}$ 
  
*$P(X=6)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}\times \frac{5}{5}=\frac{1}{252}$ 
  

My solution was


*

*$P(X=1)=\frac{5\times 9!}{10!}=\frac{1}{2}$ because there are five women to occupy 1st rank and then there remained 9 which can permute in $9!$ ways. There are total $10!$ ways to permute $10$ people

*$P(X=2)=\frac{5\times \binom{5}{4}\times 8!}{10!}=\frac{5}{18}$ because there are five women to occupy 2st rank. The 1st rank will be of a man. So we have to select $4$ out of $5$ men which will be ranked after $2$nd rank. These four men and remaining 4 women can be permuted in $8!$ ways.

*$P(X=3)=\frac{5\times \binom{5}{3}\times 7!}{10!}=\frac{5}{72}$

*$P(X=4)=\frac{5\times \binom{5}{2}\times 6!}{10!}=\frac{5}{504}$

*$P(X=5)=\frac{5\times \binom{5}{1}\times 5!}{10!}=\frac{5}{6048}$

*$P(X=6)=\frac{5\times \binom{5}{0}\times 4!}{10!}=\frac{1}{30240}$


Doubts


*

*Where my logic went wrong?

*When I compared the two approach, I realized that the books solution is permuting ranks higher than the highest ranked girl, while my solution is permuting ranks lower than the highest ranked girl. So I was guessing what makes book solution not permute lower ranks and my solution not permuting higher ranks. Shouldn't we permute on both sides of highest ranked girl?


In your calculations,


*

*$5$ is the number of possible top-ranked women

*$\binom{5}{k}$ is the number of ways $k$ of the five men can have a lower rank than the top-ranked woman

*$(9 - k)!$ is the number of ways of arranging the $9 - k$ people whose rankings are lower than that of the top-ranked woman

*$10!$ is the number of possible sequences of rankings


In your numerators, you failed to multiply by the number of ways the men who are selected before the first woman can be ranked.  Observe that 
\begin{align*}
P(X = 1) & = \frac{0! \cdot 5 \cdot \binom{5}{5} \cdot 9!}{10!} = \frac{1}{2}\\
P(X = 2) & = \frac{1! \cdot 5 \cdot \binom{5}{4} \cdot 8!}{10!} = \frac{5}{18}\\
P(X = 3) & = \frac{2! \cdot 5 \cdot \binom{5}{3} \cdot 7!}{10!} = \frac{5}{36}\\
P(X = 4) & = \frac{3! \cdot 5 \cdot \binom{5}{2} \cdot 6!}{10!} = \frac{5}{84}\\
P(X = 5) & = \frac{4! \cdot 5 \cdot \binom{5}{1} \cdot 5!}{10!} = \frac{5}{252}\\
P(X = 1) & = \frac{5! \cdot 5 \cdot \binom{5}{0} \cdot 4!}{10!} = \frac{1}{252}
\end{align*}
The reason you obtained the correct answer for $X = 1$ and $X = 2$ is that $0! = 1! = 1$.  
Note:  The author(s) of your text are calculating the probability that the top-ranked woman is in the $k$th position.  For that to occur, we must select $k - 1$ of the five men to be ranked ahead of her and one of the five women to fill the $k$th position while choosing $k$ of the $10$ people.  Hence, the answer in the text is equivalent to
\begin{align*}
P(X = 1) & = \frac{\binom{5}{0}\binom{5}{1}}{\binom{10}{1}} = \frac{1}{2}\\
P(X = 2) & = \frac{\binom{5}{1}\binom{5}{1}}{\binom{10}{2}} = \frac{5}{18}\\
P(X = 3) & = \frac{\binom{5}{2}\binom{5}{1}}{\binom{10}{3}} = \frac{5}{36}\\
P(X = 4) & = \frac{\binom{5}{3}\binom{5}{1}}{\binom{10}{4}} = \frac{5}{84}\\
P(X = 5) & = \frac{\binom{5}{4}\binom{5}{1}}{\binom{10}{5}} = \frac{5}{252}\\
P(X = 6) & = \frac{\binom{5}{5}\binom{5}{1}}{\binom{10}{6}} = \frac{1}{252}
\end{align*}  
