Kenneth Rosen Edition 6th Exercise 5.3 Question 13 (Page No. 361)

Question

I was doing two Questions of rosen counting chapter

Q1. A group contains n mens and n women. How many ways are there to "arrange" these people in a row if the men and women alternate ?

Answer is 2(n!)^2, I agree with logic and answer.

 

Q2. How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other?

Answer is P(8,8)*P(9,5), I also agree with answer and logic.

 

My dobut is, The frame of questions are similar but then why can't we solve Q1 with the approach used in Q2 i.e. first permuting n men and then choosing n women places from (n+1) places created my positioning the men and permuting them again i.e P(n,n)*P(n+1,n) ??

Comments

the first method is right because using second method alternate positioning is not guaranteed , like i can fill the five places as follows mwmw_wmwm or mw_wmwmwm ,etc

instead use 4 places instead of five leaving one end restricted to a woman like _w_w_w_w, this way you can arrange in C(4,4) ways and permute them and in the result multiply with 2 for restricting the other end like w_w_w_w_

Answer 1

First, position the $n$ men relative to each other. There are $n!$ ways to do this.
This creates $n+1$ slots where a woman may stand.
We need to choose $n$ places from these $n+1$ slots by considering the condition men and women should alternate.
So the only possible n slots are:

• In front of the first man, between the fist and second men, ..., between the $n-1$th and $n$th men.
  or
• between the first and second men, ..., between the $n-1$th and $n$th men, and behind $n$th man.

ie. $2$ ways.
After selecting the slots, we can position the women in $n!$ ways.
$\therefore$ answer is: $n!\cdot 2 \cdot n!$ = $2\cdot {n!}^2$

As you said, we can't use $P(n+1, n)$ since it violates the given condition.