Kenneth Rosen Edition 6th Exercise 5.3 Question 24 (Page No. 361)

Question

How many ways are there to arrange $6$ Men and $10$ Women in a row such that no two men are together.

This can simply be answered by putting $10$ women first which gives $11$ spaces for men and the final answer will be * $10!* \binom{11}{6} * 6!$ . But I have an alternate strategy

Let 10 women be divided into $5$ groups such that one group contains $6$ women and the rest contain single woman.

this can be done in $\binom{10}{6} * 6! * 5!$ Now we have $6$ spaces for $6$ men .So finally we have $\binom{10}{6} * 6! * 5! * 6!$

Why the both answers are not same. What am I missing? Which approach is correct?

Comments

you don't have 6 spaces for 6 men. you still have 11 places for 6 men

If you do $\binom{10}{6}$ * $6!$ * $5!$ * $\binom{11}{6}$ * 6! will give the same answer which is $10! * 11P_{6}$

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I wanted to create only 6 spaces for Men that is why I created those groups. You can put a Man only between the 5 groups not inside them, for example, M(WWWWWW)M(W)M(W)M(W)M(W)M
But I now know that this approach is faulty. You can think of positioning the Men as Dividing the Women into groups of 5. Those 5 groups may have some number of women, like 2, 2, 3, 2, 1 and a Man is put between each group.

My mistake was that I considered only one instance of the groups namely 6, 1, 1, 1, 1. But as the above example shows there can be multiple such groups.

This shows that the first approach is correct and more intuitive.

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the 2nd method will lead to different solution which is not correct

Answer 1

At first, we'll place the women

Now, there are $10$ women

$10$ women will occupy $10$ places

& they can arrange their places in $10!$ ways

We can see that, now there are $11$ vacant places.

& there are $6$ men

So, Among $11$ places $6$ men can sit in $^{11}P_6$ ways [Choose $6$ places from $11$ places & then arrange the places]

∴ Total Ways = $10! \times ^{11}P_6$ ways

$\qquad \qquad = 10! \times \dfrac{11!}{5!}$

$\qquad \qquad = 1,20,708,40,32,000$ ways. $\rightarrow$ This is the answer.

But, you want to do this in another way -

which is to divide the women into $5$ groups, where $1$ group will contain $6$ women, & the remaining $4$ groups will contain $1$ women each.

So, let's assume $1^{st}$ group will having $6$ women 

$2^{nd}$ group will contain $1$ women

$3^{rd}$ group will contain $1$ women

$4^{th}$ group will contain $1$ women

$5^{th}$ group will contain $1$ women 

Now, $5$ groups can arrange in $5!$ ways

from $10$ women we've to choose $6$ women who'll be in $1^{st}$ group, that can be done in $^{10}C_6$ ways

& the group ($1^{st}$ group ) which has $6$ women, in that group $6$ women can arrange their places in $6!$ ways

The scenario will look like-

Now, vacant places are-

 = $11$

∴ $6$ men can sit in $11$ places in $^{11}P_6$ ways [first, they'll choose $6$ places from $11$ places & then they arranged their position.]

∴ Total Ways = $5! \times ^{10}C_6 \times 6! \times ^{11}P_6$

$\qquad \qquad = 5! \times 210 \times 6! \times ^{11}P_6 $

$\qquad \qquad = 6.03542016e12$

But we'll not use this method to solve this problem.

It'll give us a different solution as in the question there is not mentioned anything whether women are in group or not

& when we divide 10 women into 5 groups, it will increase the total no. ways by $5$ times

[$1207084032000 \times 5 = 6.03542016e12$]