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?


I would like to suggest another way of getting the results. For me this way is "combinatorically more intuitive":


*

*If $X = i$, then there are $10-i$ ranks left for the remaining $4$ women. So. there are $\binom{10-i}{4}$ ways of choosing 4 further ranks.

*As each permutation of the $5$ men and $5$ women gives another way of ranking we need to multiply by $5! \cdot 5!$

*All together (it gives exactly the values from the solution presented): $$P(X = i) = \frac{\binom{10-i}{4}\cdot5! \cdot 5!}{10!} =  \frac{\binom{10-i}{4}}{252} \mbox{ for } i = 1,...,6$$

