$m$ men and $n$ women in a roundtable

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The question has been discussed before here.



The answer one of the respondents has given is $(m-1)! cdot n! cdot C(m,n)$.



I found the question solved in here.



Where the author has given the answer as (for no $2$ men sitting together):



$$m! cdot n! cdot [C(n-1,m-1) + C(n,m)]$$



I could follow his argument but could not understand why he didn't use $(n-1)!$ instead of $n!$ since we have circular permutations.



Here is the soln in the book.
Solution




combinatorics




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  • Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Aug 22 at 10:25










  • Sure. Thanks. Will do
    – sidharth
    Aug 22 at 10:34






  • 2




    @sidharth, Can you please upload the solution from the book?
    – prog_SAHIL
    Aug 22 at 11:23










  • I have added the solution
    – sidharth
    Aug 23 at 2:41










  • My 2 cents... and I'd love for someone else to post a proper answer. I think your specific question is why is it $m!n!$ and not $(m-1)!(n-1)!$ Generally you want to use $(n-1)!$ for circles because you don't want to double count arrangements. But in the solution the dude changes this from a circular arrangement problem to a line arrangement problem by fixing on a specific chair $X$ and looking at the two cases where a man is in the first chair of the line or a girl is. It is now a line arrangement problem. The solution has two components, one is figuring out where guys and girls can sit.
    – HJ_beginner
    Aug 23 at 5:20















up vote
0
down vote

favorite












The question has been discussed before here.



The answer one of the respondents has given is $(m-1)! cdot n! cdot C(m,n)$.



I found the question solved in here.



Where the author has given the answer as (for no $2$ men sitting together):



$$m! cdot n! cdot [C(n-1,m-1) + C(n,m)]$$



I could follow his argument but could not understand why he didn't use $(n-1)!$ instead of $n!$ since we have circular permutations.



Here is the soln in the book.
Solution




combinatorics




share|cite|improve this question






















  • Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Aug 22 at 10:25










  • Sure. Thanks. Will do
    – sidharth
    Aug 22 at 10:34






  • 2




    @sidharth, Can you please upload the solution from the book?
    – prog_SAHIL
    Aug 22 at 11:23










  • I have added the solution
    – sidharth
    Aug 23 at 2:41










  • My 2 cents... and I'd love for someone else to post a proper answer. I think your specific question is why is it $m!n!$ and not $(m-1)!(n-1)!$ Generally you want to use $(n-1)!$ for circles because you don't want to double count arrangements. But in the solution the dude changes this from a circular arrangement problem to a line arrangement problem by fixing on a specific chair $X$ and looking at the two cases where a man is in the first chair of the line or a girl is. It is now a line arrangement problem. The solution has two components, one is figuring out where guys and girls can sit.
    – HJ_beginner
    Aug 23 at 5:20













up vote
0
down vote

favorite









up vote
0
down vote

favorite











The question has been discussed before here.



The answer one of the respondents has given is $(m-1)! cdot n! cdot C(m,n)$.



I found the question solved in here.



Where the author has given the answer as (for no $2$ men sitting together):



$$m! cdot n! cdot [C(n-1,m-1) + C(n,m)]$$



I could follow his argument but could not understand why he didn't use $(n-1)!$ instead of $n!$ since we have circular permutations.



Here is the soln in the book.
Solution




combinatorics




share|cite|improve this question














The question has been discussed before here.



The answer one of the respondents has given is $(m-1)! cdot n! cdot C(m,n)$.



I found the question solved in here.



Where the author has given the answer as (for no $2$ men sitting together):



$$m! cdot n! cdot [C(n-1,m-1) + C(n,m)]$$



I could follow his argument but could not understand why he didn't use $(n-1)!$ instead of $n!$ since we have circular permutations.



Here is the soln in the book.
Solution





combinatorics





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 at 4:53

























asked Aug 22 at 10:15









sidharth

93




93











  • Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Aug 22 at 10:25










  • Sure. Thanks. Will do
    – sidharth
    Aug 22 at 10:34






  • 2




    @sidharth, Can you please upload the solution from the book?
    – prog_SAHIL
    Aug 22 at 11:23










  • I have added the solution
    – sidharth
    Aug 23 at 2:41










  • My 2 cents... and I'd love for someone else to post a proper answer. I think your specific question is why is it $m!n!$ and not $(m-1)!(n-1)!$ Generally you want to use $(n-1)!$ for circles because you don't want to double count arrangements. But in the solution the dude changes this from a circular arrangement problem to a line arrangement problem by fixing on a specific chair $X$ and looking at the two cases where a man is in the first chair of the line or a girl is. It is now a line arrangement problem. The solution has two components, one is figuring out where guys and girls can sit.
    – HJ_beginner
    Aug 23 at 5:20

















  • Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Aug 22 at 10:25










  • Sure. Thanks. Will do
    – sidharth
    Aug 22 at 10:34






  • 2




    @sidharth, Can you please upload the solution from the book?
    – prog_SAHIL
    Aug 22 at 11:23










  • I have added the solution
    – sidharth
    Aug 23 at 2:41










  • My 2 cents... and I'd love for someone else to post a proper answer. I think your specific question is why is it $m!n!$ and not $(m-1)!(n-1)!$ Generally you want to use $(n-1)!$ for circles because you don't want to double count arrangements. But in the solution the dude changes this from a circular arrangement problem to a line arrangement problem by fixing on a specific chair $X$ and looking at the two cases where a man is in the first chair of the line or a girl is. It is now a line arrangement problem. The solution has two components, one is figuring out where guys and girls can sit.
    – HJ_beginner
    Aug 23 at 5:20
















Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 22 at 10:25




Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 22 at 10:25












Sure. Thanks. Will do
– sidharth
Aug 22 at 10:34




Sure. Thanks. Will do
– sidharth
Aug 22 at 10:34




2




2




@sidharth, Can you please upload the solution from the book?
– prog_SAHIL
Aug 22 at 11:23




@sidharth, Can you please upload the solution from the book?
– prog_SAHIL
Aug 22 at 11:23












I have added the solution
– sidharth
Aug 23 at 2:41




I have added the solution
– sidharth
Aug 23 at 2:41












My 2 cents... and I'd love for someone else to post a proper answer. I think your specific question is why is it $m!n!$ and not $(m-1)!(n-1)!$ Generally you want to use $(n-1)!$ for circles because you don't want to double count arrangements. But in the solution the dude changes this from a circular arrangement problem to a line arrangement problem by fixing on a specific chair $X$ and looking at the two cases where a man is in the first chair of the line or a girl is. It is now a line arrangement problem. The solution has two components, one is figuring out where guys and girls can sit.
– HJ_beginner
Aug 23 at 5:20





My 2 cents... and I'd love for someone else to post a proper answer. I think your specific question is why is it $m!n!$ and not $(m-1)!(n-1)!$ Generally you want to use $(n-1)!$ for circles because you don't want to double count arrangements. But in the solution the dude changes this from a circular arrangement problem to a line arrangement problem by fixing on a specific chair $X$ and looking at the two cases where a man is in the first chair of the line or a girl is. It is now a line arrangement problem. The solution has two components, one is figuring out where guys and girls can sit.
– HJ_beginner
Aug 23 at 5:20
















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