How many ways are there in the case that there is ONLY ONE adjacent couple?

Question

"Three couple will be arranged in a row.How many ways are there in the case that there is ONLY ONE couple which are adjacent."

For example , let $(a_1,a_2),(b_1,b_2),(c_1,c_2)$ be the couples, so they cannot be arranged in the form of $a_1,c_1,c_2,b_2,b_1,a_2$. There can be only one couple adjacent such as $b_1,a_1,a_2,c_1,b_2,c_2$ is acceptable.

I found this question in my textbook.I tried that $C(3,1).5!.2!-[C(3,2).6.2!.2!.2!+C(3,3).2!.2!.2!.3!]$ but my answer is wrong and I am confused. Can you help me?

Answer 1

There are three ways to select the adjacent couple. We then have five objects to permute, the couple and the other four people. The five objects can be arranged in $5!$ ways. The members of the couple can be arranged internally in $2!$ ways. Hence, there are $$\binom{3}{1}5!2!$$ arrangements in which a couple is together.

However, we have counted each case in which two couples in which the members of a couple are adjacent twice, once for each way of designating one of the couples as the couple that is together. We do not want to count such cases at all, so we must subtract twice the number of cases in which there are two couples in which the members of a couple are adjacent.

There are $\binom{3}{2}$ ways to select two couples in which the members of a couple are adjacent. That gives us four objects to arrange, the two couples and the two individuals from the third couple. The objects can be arranged in $4!$ ways. Each adjacent couple can be arranged internally in $2!$ ways. Hence, there are $$\binom{3}{2}4!2!2!$$ such cases.

That gives us a running total of $$\binom{3}{1}5!2! - 2\binom{3}{2}4!2!2!$$ but we have not yet accounted for those arrangements in which there are three couples in which the members of the couple are adjacent to each other.

We counted the cases in which there are three couples are together three times when we counted arrangements in which the members of a couple are together, once for each way of designating one of the three couples as the couple with adjacent members. Observe that we counted cases in which there are three couples with adjacent members three times when we counted cases in which there are two couples with adjacent members, once for each of the $\binom{3}{2}$ ways of designating two of the three couples as the couples which are together. Since we subtracted twice the number of arrangements in which there are two couples with adjacent members, we subtracted cases in which there are three couples with adjacent members six times. We don't want to count such arrangements at all, so we need to add three times the number of arrangements in which there are three couples with adjacent members.

If there are three couples with adjacent members, there are $3!$ ways to arrange the couples and $2!$ ways to arrange each couple internally. Hence, there are $$3!2!2!2!$$ such arrangements.

Consequently, by the generalized Inclusion-Exclusion Principle, there are $$\binom{3}{1}5!2! - 2\binom{3}{2}4!2!2! + 3\binom{3}{3}3!2!2!2!$$ arrangements of three couples in which there is only one adjacent couple.

Answer 2

Suppose the couples are $A,B,C$. Since exactly one couple is to be together, the number will be $3$ times the number only couple $A$ is together.

To apply inclusion-exclusion, we shall have to exclude the $2$ ways in which two couples including $A$ can be formed and be together and add one formation in which all three couples are together.

For the computation, it is convenient to consider couples who are together as "individuals" for this purpose.

Thus we get $\;3[2^1\cdot5! - 2\cdot2^2\cdot4! +1.2^3\cdot 3!] = 288$

Added explanation for first term

$2^1$ because a couple can sit in $2$ ways, and $5!$ gives permutations of $4+1$ "individual"