Probability help!

Here's the answer....although someone else can probably present it in a more straightforward way........!!!!

Seat Thomas (T) in chair 1  and Lily (L) in any chair from 3 - 7....she has 5 choices

Richard/Hang (RH) can be considered as one entity and can be arranged in 2 ways each

And the other 4 people can be arranged in 4! =  24 ways.....so we have

T 2 L 4 5 6 7 8 ... RH (4 choices  x 2 arrangements each )  x  24  = 192 x 5 choices for Lily = 960 

 If  Thomas is seated in chair 1 and Lily in chair 8 , we have

T 2 3 4 5 6 7 L ...RH (5 choices x 2 arrangements) x 24  = 240

So....when Thomas is seated in chair 1 there are

960 + 240    = 1200  arrangements possible

Next ....seat Thomas in chair 2  and Lily in any of the chairs 4 - 7....she has 4 choices

Following the above notations, we have

1 T 3 4 5 L 7 8    ...RH (3 choices  x 2 arrangements) x  24  = 144 x 4 choices for Lily  = 576

And when Thomas is seated in chair 2 and Lily in chair 8 we have

1 T 3 4 5 6 7 L  ...  RH 4 choices x 2 arrangements X 24 = 192

So....when Thomas is seated in chair 2 there are

576 + 192  = 768  arrangements possible

When Thomas is seated in chair 3, we have the following :

L 2 T 4 5 6 7 8 ...  RH  (4 choices  x  2 arrangements)  x 24   = 192

1 2 T 4 [Lily can occupy any chair 5 - 7]  8   ...  RH (3 choices x 2 arrangements) x 24 x 3 choices for  Lily = 144 x 3 

1 2 T 4 5 6 7 L ....   RH (4 x 2) x 24 =  192

So when Thomas is seated in the third chair we have

192*2 + 144*3  = 816  arrangements possible

And this number of arrangements wll be true when Thomas occupies any of the chairs 3 - 6

So....we have 816 (4) = 3264 possible arrangements when Thomas occupies chairs 3 - 6

When Thomas occupies chair 7, there will be the same number of arrangements as when he occupies chair 2  = 768

And when he occupies chair 8, there will be the same number of arrangements as when he occupies the first chair  = 1200

So we have

1200(2) + 768(2) + 816(4)  = 7200 total arrangements

Thanks Chris

How many ways can richard and hang sit together.

Put a lasso around them. It could be Richard then Hang or hang then Richard but now there are 7 entities. That is 7! ×2 ways.

Now how many ways can (Hang and Richard) and (Thomas and lily) sit together.

Now there are 6 entities so that is 6!×2×2=4×6!

So the number of ways that Richard and hang Do sit together. AND Thomas and Lily DO NOT sit together is

2×7! - 4×6! = 7200

Just like Chris already determined :)