We have to seat 10 people in a row.
Condition: two people always sit together and two people never sit together.
My attempt: Let the two people who always sit together be taken as 1 person for the time being. So, we have nine people.
1) Keep one of two people who never sit together at seat 1. Then, we have $7\cdot8!$ ways to seat others.
2) Keep that person at seat 2. Then, we have $6\cdot8!$ ways to seat others.
3) Adding up the above iterations, we have $8!(7\cdot2+6\cdot7)$ ways.
4) Finally the two pairs can be permutated in 2 ways each.
Hence, the answer should be $4\cdot8!(7\cdot2+6\cdot7)$.
Is this correct?
Is there a better approach to solve this problem ? Please advise.