# Help I Dont Know How To Do This [factorial Question]

• Sep 11th 2007, 01:52 PM
imarc
Help I Dont Know How To Do This [factorial Question]
There are 10 students in a rectangular table. Tanya has to sit at the head table. Henry cannot sit beside Nancy or Wilson. How many ways can you rearrange the sitting?

I cant do this question can someone help!!!

Thanks so much
Marc
• Sep 11th 2007, 02:12 PM
Jhevon
Quote:

Originally Posted by imarc
There are 10 students in a rectangular table. Tanya has to sit at the head table. Henry cannot sit beside Nancy or Wilson. How many ways can you rearrange the sitting?

I cant do this question can someone help!!!

Thanks so much
Marc

is the arrangement such that one person can sit at the head, one at the end, and 4 on each side?
• Sep 11th 2007, 02:18 PM
imarc
yes
• Sep 11th 2007, 03:11 PM
Plato
Once we have established the “head of table”, the table is now ordered. All we need to do is count the number of ways to seat the other nine. That is (9!) ways without any other restrictions. However, there are other restrictions. So lets count the cases we don’t want. There are 2(8!) ways for Nancy and Henry are seated together. The same for Henry and Wilson to be seated together. But if we add those two we have counted the cases where Nancy, Henry, and Wilson are seated as a threesome with Henery between them: 2(7!).

Remove the ones we don’t want from the total:
(9!)-[2(8!)+2(8!)-2(7!)].
• Sep 12th 2007, 03:40 AM
imarc
Quote:

Originally Posted by Plato
Once we have established the “head of table”, the table is now ordered. All we need to do is count the number of ways to seat the other nine. That is (9!) ways without any other restrictions. However, there are other restrictions. So lets count the cases we don’t want. There are 2(8!) ways for Nancy and Henry are seated together. The same for Henry and Wilson to be seated together. But if we add those two we have counted the cases where Nancy, Henry, and Wilson are seated as a threesome with Henery between them: 2(7!).

Remove the ones we don’t want from the total:
(9!)-[2(8!)+2(8!)-2(7!)].

i understand where you got the 9! but howdid you get the2 and (8!)
• Sep 12th 2007, 05:24 AM
Plato
To count the number of ways for Henry and Nancy may sit together; think of them as one unit. Now instead of nine to arrange we have eight, (8!). But we can have HN or NH in those orders thus 2(8!) is the number of ways Henry and Nancy may sit together. For all three we have NHW or WHN which counts a one block thus 2(7!).
• Sep 12th 2007, 07:48 AM
ThePerfectHacker
There is a nice application of Burnside's formula on G-sets which I learned few years ago to doing these rotational combinatorics problem. Here. (post #14).