Then consider the adjacent situation: n couples, (considering the whole couple and the exchange position of husband and wife), choose a couple, there are n options, you can put them in any of the n positions, there are n release methods, and the remaining 2n-2 people are randomly arranged, so the adjacent arrangement is 2*n*n*(2n-2)! This situation;

So the total arrangement of * * * is (2n)! -2*n*n*(2n-2)！

=2n*(2n- 1)*(2n-2)！ -2*n*n*(2n-2)！

=2n*(n- 1)*(2n-2)！ Kindness

For example: n=2, let AB and CD be two pairs, then the possible arrangement is as follows.

ACBD、ADBC、BCAD、BDAC； Cadb, dacb, cbda and dbca.

2*2*(2- 1)*(2*2-2)! =2*2* 1*2! =8。

Ha ha. It looks right.

Or two couples with four people, * * * with four! =24 arrangements,

Adjacent AB, ABCD, ABCD; DABC CABD； CDAB and DCAB；;

There are also six adjacent BA's, namely the exchange positions on AB, BACD and BADC；; ; DBAC CBAD； DCBA CDBA；

There are six kinds of adjacent CDs and DC, some of which have appeared above, but none of them appear at the same time:

ACDB，ADCB，BCDA，BDCA。

Therefore, there are 6+6+4 adjacent permutations and 24- 16=8 non-adjacent permutations.