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

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- September 11th 2007, 01:52 PMimarcHelp 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 - September 11th 2007, 02:12 PMJhevon
- September 11th 2007, 02:18 PMimarc
yes

- September 11th 2007, 03:11 PMPlato
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!)]. - September 12th 2007, 03:40 AMimarc
- September 12th 2007, 05:24 AMPlato
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!).

- September 12th 2007, 07:48 AMThePerfectHacker
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).