My solution is:
- Choose $1$ man from $6$ men: $6$ ways
- Choose $1$ woman from $6$ women: $6$ ways
- Arrange the rest of the people: $10!$ ways
- Insert the pair in the arranged line: $11$ ways
- Swap a man and a woman in the pair: $2$ ways
Therefore, the result is $6 \cdot 6 \cdot 10! \cdot 11 \cdot 2 = 11! \cdot 72$
But my friends argue that we don't have to choose a man and a woman in the pair because the problem already told us that there is just $1$ man and $1$ woman that must stand next to each other. My argument is that if we don't choose which pair it is, we can't write the outcomes because we don't know which pair it is.
Which one is the right way?
Your friends have interpreted the question correctly. The particular man and woman who stand next to each other are given. Therefore, the first two steps of your proposed solution are superfluous. The final three steps, which you did correctly, yield the answer $10! \cdot 11 \cdot 2 = 2 \cdot 11!$.
Alternate Solution: Treat the particular man and woman who must stand next to each other as a single object. Then we have eleven objects to arrange, a block containing the man and woman in question and the other ten people. The eleven objects can be arranged in $11!$ ways. The man and woman can be arranged within the block in $2!$ ways. Hence, there are $11!2!$ admissible arrangements.