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Six married couple are to be seated at a circular table. In how many ways can they arrange themselves so that no wife sits next to her husband? 0 0 comment dev

Let's first count number of ways in which at least one husband and his wife sit next to each other. For this, group one husband and wife, so that they site next to each other. Consider this group and other $10$ persons, together as $11$ persons. These $11$ persons can be arranged in $(11-1)!$ ways. The husband and wife, can interchange their positions in the group. So we need to multiply by $2!.$ Since we can group any of the $6$ couples, further we need to further multiply with $6\text{C}1.$ Therefore, number of ways in which at least one husband and his wife sit next to each other is $(11-1)!(2!)6\text{C}1$

But we have over-counted the above. To get the correct count, apply inclusion-exclusion principle. (This concept will be better understood with a small set)

Group two couples. Take the two groups and remaining $8$ persons as $10$ persons. They can be arranged in $(10-1)!$ ways. The husband and wife, can interchange their positions in each of the two groups $(2!)^2$ and total $6\text{C}2$ such groups. That is, total $(10-1)!(2!)^{2}6\text{C}2$ such arrangements.

Continue the same with $3,4,5,6$ groups and apply inclusion-exclusion principle.

So we get, number of ways in which at least one husband and his wife sit next to each other as
$(11-1)!(2!)6\text{C}1-(10-1)!(2!)^{2}6\text{C}2+(9-1)!(2!)^{3}6\text{C}3-(8-1)!(2!)^{4}6\text{C}4+(7-1)!(2!)^{5}6\text{C}5-(6-1)!(2!)^{6}6\text{C}6=27144960$

We know, number of ways in which $12$ persons can be seated around a circular table is $11!$

Therefore, number of arrangements where no wife sits next to her husband $=11!-27144960=12771840$

See this question for different cases coming under the same problem. 0 0 comment