There is six couples - husband and wives in each couple. They sit in a row of 12 seats. In how many different ways can the 12 people sit so that the 6 wives all sit next to each other, and none of the wives sits next to her own husband.
After reading the answer and thinking thoroughly, i still am confused over one part: Let me present you my own solution:
Case 1: The 6 wives all sit at either corners: In which the ways : $$6! \times {5 \choose 1} \times 5! \times 2 = 864000$$
Case 2:
Subcase 1: Left neighbor of wife is husband of the rightmost wife.
We have $${5 \choose 1} \times 5! \times 6! = 432000$$ since we do not need to care about the fixed husband (who is the husband of the rightmost wife).
Subcase 2: Left neighbor of wife is not husband of the rightmost wife:
This is the part where i differ from the answer: I think it should be $$6! \times {4 \choose 1} \times {4 \choose 1} \times 4!$$
However the answer for subcase 2 is $$6! \times {4 \choose 1} \times {4 \choose 1} \times 5!$$
May i know why? My train of thoughts is since we already chosen the 2 husbands above, we just need to deal with the remaining 4 permutations.
Edit: Now i know why they times $5!$ because you need to consider 5 different scenarios.