There are 3 married couples should sit in a straight row of 6 seats. If couples sit random, what is the chance that at least one man sits next to his wife?
And, what is the chance that exactly $3$ ($2$, and $1$ also) couples are thogether?
Total number of combinations: $6!=720$
I want to counting $3$ disjoints sets.
$A_1$: the number of combinations in which there are exactly 3 men next to his wifes, denoting each couple by $P_i$ we have $3!$ possible combinations $P_1 P_2 P_3, P_2P_3P_1,$ etc.. But each couple can order in $2!$ forms then:
$A_2$: the number of combinations in which there are exactly 2 men next to his wife, denoting each couple by $P_i$ we have $4!-3$ possible combinations if we fix $2$ couples $P_1, P_2$, since there are $2!3$ of $4!$ combinations in which the third couple is not togheter namely $P_1 a P_2 b$, $a P_1 b P_2$ and $a P_1 P_2 b$ and changing $P_1$ with $P_2$. And since each couple can ordered in $2!$ forms we need to add factors $2!2!2!$ like previous case. Then
$A_2=2!2!2!3(4!-2!3)=432$ But im not sure!!!
There is an error in $A_2=2!2!2!3(4!-2!3)=432$, since $2!2!3$ of $4!$ was favorable cases. Then: $A_2=2!2!3(2!2!3)=2!^43^2=144$ as noted Brian.
$A_3$: the number of combinations in which there are exactly 1 man next to his wife. I have not idea how to compute this monster.
$A_3=288$ solved by Brian. $A_1+A_2+A_3=480$