help me solve this stats/probability question

Question

Eight 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 different ways can the people be seated for dinner? Show all your work.

Answer 1

Break this into two parts.

There are 2 seats that Henry can sit in next to Tanya and 5 seats he can sit in in which he is not next to Tanya.

In the 2 seats that Henry sits next to Tanya, Wilson and Nancy can sit anywhere except in Tanya's seat, Henry's seat, and next to Henry, so there are 8-3=5 seats available, and 5*4 ways Wilson and Nancy can sit.

Thus, the total number of ways is 2 * 5 * 4

In the 5 seats that Henry sits not next to Tanya, Wilson and Nancy can sit anywhere except in Tanya's seat, Henry's seat, and the two seats on either side of Henry, so there are 8 - 4 = 4 seats available, so there are 4*3 ways that Wilson and Nancy can sit.

Thus, the total number of ways is 5 * 4 * 3

Once Tanya, Henry, Wilson, and Nancy are seated, there are 4 seats left for the 4 remaining people, so they can sit in 4! ways.

Then,the total number of ways that the people can be seated is

(2*5*4+5*4*3)*4! = (5*4*(2+3))*4! = 5²*4*4! = 2400