Ten people are to be seated at a rectangular table for dinne

We don't have to do this question, but I'd liketo know how to do this:

Ten people are to be seated at a rectangular table for dinner. Tanya will sit at the head of the table. Henry must not sit beside either Wilson or Nancy. In how many ways can the people be seated for dinner?

So from what you guys have taught me, For the Henry part, you would do 8!*4? = 161 280
then Ooo so I got it now, its 9!- (8!*4)= 201 600 , but since I took tanya out of it..is that right? I did 9!, isnt that like excluded Tanya totally? or since shes just not moving,it means she can't be mixed up with the others?

I wonder if this problem is considering only 1 head of the table.

Anyway, we could count the number of ways they can sit together and subtract from 9!.

Sit Tanya at the head and arrange the others 9! ways.

If we tie Henry, Wilson and Nancy together, there are 7! ways to seat them all. Also, if all 3 are seated next to one another, we can arrange them in 3! ways. If Henry sits beside Wilson, we can arrange in 2! ways.
If Henry sits beside Nancy, in 2! ways.

\(\displaystyle \L\\\frac{7!}{3!2!2!}=210\)

6! to arrange the other 6 people.

9!-P(7,3)6!=211680

Multiply by 2 if there are 2 heads of the table.

Did I count right?. Maybe pka, the Count, Soroban.mark, etc. may be along to confirm or deny.

I agree with Galactus's answer. I had a different approach, which is similar to Anna's, so it might be easier to follow, who knows...

9! - 4*8! is a good approach, but it subtracts the cases when Henry sits next to both of them twice! So I added that...

9! - 4*8! + 2*7! = 211680 as well.

Only way Henry can sit next to both of Wilson and Nancy is when he is in the center, that's why I multiplied the last by 2. You may also see the reason for adding that from the formula n(AUB) = n(A)+n(B)-n(A intersection B).

Let's see what a fellow Ontarian can do!

Since Tanya she no moves there, then we really have a simpler problem:
how many ways can the digits 1 to 9 be arranged such that 12, 21, 13, 31
are not in the lineup? 1=Henry, 2=Wilson, 3=Nancy.

No restriction: 9! = 362880
Restrictions: 8*4*7! = -161280
Add triplets: 2 *7! = 10080 : triplets like 213 and 312
Leaves: 211, 680

Oui?

Edit: looks like I matched the 2 pro's :wink: