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.
Best Answer
The way I would solve this is taking the number of ways of seating the ten people without the condition that the two don't sit together ($2 \cdot 9!$) and subtracting the number of configurations where they sit together ($2^2 \cdot 8!$). So the answer is $2 \cdot 9! - 2^2 \cdot 8! = 564480$.
Your method would also work, but you have made two mistakes. Firstly your $8!$'s should be $7!$, we are seating nine people in total and the two who don't sit together have already been seated, so there are seven people left. Secondly, you should not multiply the final answer by $2$ for the people who don't sit together, you have already counted both ways for them to sit. So the correct answer is $2 \cdot 7!(7 \cdot 2 + 6 \cdot 7) = 564480$.