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.