Dance couples permutation if one particular girl will not dance with one particular boy

Question

I see there were similar questions but I need help and advice with this one.

So, there are $12$ boys and $12$ girls. How many couples we can create if one couple is arguing and won't dance together?

I know that without restrictions we can create couples in $12!$ ways, but with one fighting, is it correct to presume that answer is $11!$ because sooner or later they could find themselves in combination?

And second part is if they dance in circle (boy-girl-boy-girl-...), in how many ways they can create a circle if one couple is arguing and won't dance next to each other?

We use formula for "round table" so first we decide where is first boy's position and then without restrictions we have $11!$ possible ways to create circle without girls and then girls take their position in-between boys with $12!$ ways (so without restrictions answer would be $11! \cdot 12!$), but how can I take in account couple in fight refusing to dance next to each other? Or my approach is wrong and I should use the principle of inclusion-exclusion?

Please help and tnx!

Answer 1

In how many ways can $12$ boys and $12$ girls form opposite sex couples for dancing if a particular boy and girl will not dance together?

Suppose the girls choose their dance partners. We allow the girl who is arguing with a boy to choose first. Since she will not choose him, she has eleven options. Once she has made her choice, the remaining boys can be matched with the remaining girls in $11!$ ways. Hence, there are $11\cdot 11!$ ways opposite sex couples can be formed if that particular boy and girl do not dance together.

In how many ways can $12$ boys and $12$ girls dance in a circle with boys alternating with girls if one boy and girl will not dance next to each other.

Suppose the girl is Anna and the boy is Bradley. Place Anna. That determines the relative positions of the boys and girls. The girls can be placed in $11!$ ways relative to her as we proceed clockwise around the circle from Anna. This creates $12$ spaces for the boys. Bradley cannot occupy the two spaces adjacent to Anna, so he can be placed in $10$ ways. The remaining $11$ boys can fill the remaining $11$ spaces in $11!$ ways as we proceed clockwise around the circle from Anna. Hence, there are $11! \cdot 10 \cdot 11!$ admissible arrangements.

Answer 2

The answer for the first one is $11*11!$. Here is how this number comes up:

Let $G1$(girl 1) be the girl who has fought with $B1$(boy1). The girls chose which boy they want to dance with. $G1$ chooses first, $G2$ second and so on.

$G1$ has $11$ boys to choose from($12$ minus $B1$).

$G2$ has $11$ boys to choose from.

...

$G12$ "chooses" the boy that is left.

So the total number of ways they can form dance couples is $11*11*10*9*...*1=11*11!$

Another way to understand why this is the answer is that in a random assignment of boys to girls, the couple who has fought with each other will dance together with probability $11/12$, and since all assignments of boys to girls have the same probability, the number you are looking for will be $12!*\frac{11}{12}=11*11!$

For the second part of the question:

If you place $G1$ first, then there are $11!$ ways to place the other girls.

Then there are $10$ spots to place $B1$.

And then $11!$ ways to place the other $11$ boys.

So the answer should be $11!*10*11!$