# Correct logic of permuting 5 men and 5 women to find probability of different highest women rank

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

1. Where my logic went wrong?
2. 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?
• For $X=3$, done your way, sort of (but the book's solution is simpler), it should be $$\frac{(5)(4)(5)7!}{10!}$$ – quasi Aug 19 '17 at 10:36
• Also, for $X=2$, the factor $\binom{5}{4}$ is weird. It should be just $5$. Of course, the values match, but the logic is not clear, especially considering the factor $\binom{5}{3}$ in your calculation for the case $X=2$. – quasi Aug 19 '17 at 10:38
• The book only considers the people positioned before yours because they are the ones that determine if yours is the first female in the ranking. – 57Jimmy Aug 19 '17 at 10:40
• I think the problem might be that in the "big" permutations (10!, 9!,... at the end of your computations) you don't distinguish men and women. I think that the reasoning is faulty from the beginning, but by chance the first two results are correct. – 57Jimmy Aug 19 '17 at 10:48
• @57Jimmy The issue is that the OP did not consider the order of the men who finished ahead of the top-ranked woman. – N. F. Taussig Aug 19 '17 at 13:05

• $5$ 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*}

• I was guessing why book solution did not needed to consider the order of the men and women who finished after the top ranked woman. What structural difference the book solution approach have from my solution that made it to not to consider them. Is it because that the books solution tries to find the probability in each case which is affected only by the probability of higher ranks, whereas I am trying to find all possible arrangements which requires to consider permutations of people on both sides of highest ranked woman? – anir Aug 19 '17 at 14:00
• Yes. The reason you need to consider the ordering of the men before the first woman in your approach is that your sample space is the set of all possible sequences. The authors of your text opted to find the probability that the top-ranked woman is in the $k$th position, so they had to multiply the probability that she is not in the first $k - 1$ positions by the probability that she is in the $k$th position. – N. F. Taussig Aug 19 '17 at 14:25

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

1. 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.
2. As each permutation of the $5$ men and $5$ women gives another way of ranking we need to multiply by $5! \cdot 5!$
3. All together (it gives exactly the values from the solution presented): $$P(X = i) = \frac{\binom{10-i}{4}\cdot5! \cdot 5!}{10!} = \frac{\binom{10-i}{4}}{252} \mbox{ for } i = 1,...,6$$