# 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.

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.

• What does "the $6$ wives sit together" mean? Sounds like you mean a block of $6$ consecutive wives, no? – lulu Mar 20 '18 at 12:22
• In the question, i think the author meant the 6 wives sit in a block of 6 together... – nan Mar 20 '18 at 12:23
• Then I agree with your expression. I've assigned acceptable neighbors to border the wife group and I have four men left to assign at will. – lulu Mar 20 '18 at 12:29
• There does need to be a factor $5$ in there which is the number of ways to position the block of 6 wives within the row that is not at one of the ends. This needs to be included in the case 2 subcase 1 part as well. – Jaap Scherphuis Mar 20 '18 at 13:26
• @JaapScherphuis Thanks for the headsup, i realized my mistake. – nan Mar 20 '18 at 14:12

Focusing on the case in which the wives do not sit on the ends: the pattern has to be $H^aW^6H^b$ where $a,b$ are integers $≥1$ and $a+b=6$. There are clearly $5$ such patterns as there are $5$ choices for $a$, say. Fixing a pattern, your formula gives the number of ways to populate it. To get to the final count you then must multiply by $5$. I think that reconciles the apparent gap between the two formulas.