Current location - Training Enrollment Network - Mathematics courses - This question is very difficult. A math expert came in: n couples lined up to take pictures, asking that no couples were adjacent. How many kinds of queuing methods are there?
This question is very difficult. A math expert came in: n couples lined up to take pictures, asking that no couples were adjacent. How many kinds of queuing methods are there?
Firstly, there are 2n positions, which are randomly arranged as follows: (2n)! This situation;

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.