Probability that exactly r pairs of couples must sit together If n pairs of couples are seated randomly in a row, then show that the probability that exactly r pairs of couples must sit together is:
$${}_n \mathrm{ C }_r \frac{2^r(2n-r)!}{(2n)!}\sum_{k=0}^{n-r}(-1)^k{}_{n-r} \mathrm{ C }_k\frac{2^k(2n-r-k)!}{(2n-r)!},$$
where $1\le r \le n$.
If you are interested in my current approach:
Let $E_i, i=1,2,...,n $ denote the event that the $i$th couple sits together.
There are ${}_n \mathrm{ C }_r$ ways that exactly r couple sit together.
P(exactly r pairs of couples must sit together)
= ${}_n \mathrm{ C }_r P(E_1\cap E_2\cap ... \cap E_r \cap E^c_{r+1}...\cap E^c_n)$ = ${}_n \mathrm{ C }_r P((E^c_{r+1}\cap...\cap E^c_n) | E_1 \cap E_2 \cap...\cap E_r)P(E_1 \cap E_2 \cap...\cap E_r)$
= ${}_n \mathrm{ C }_r P((E_{r+1}\cup...\cup E_n)^c | E_1 \cap E_2 \cap...\cap E_r)P(E_1 \cap E_2 \cap...\cap E_r)$ = ${}_n \mathrm{ C }_r [1-P((E_{r+1}\cup...\cup E_n) | E_1 \cap E_2 \cap...\cap E_r)]P(E_1 \cap E_2 \cap...\cap E_r)$
I couldn't go further. Please help me. Thank you!
 A: First, designate $r$ of out the $n$ couples you wish to seat together. Note there are ${n \choose r}$ ways to do this.
Let $A_i$ be the event that the $i^{\text{th}}$ couple sits together. Put $A=A_1 \cap \dots \cap A_r$ and $B=A_{r+1}\cup \dots \cup A_n$. The probability that the first $r$ couples are together and the last $n-r$ souples aren't together is $$$\begin{eqnarray*}\mathbb{P}\left(A_1 \cap \dots \cap A_r \cap A_{r+1}^C\cap \dots \cap A_n^C\right) &=& \mathbb{P}(A \cap B^C) \\ &=&\mathbb{P}(A)-\mathbb{P}(A\cap B) \\ &=& \frac{1}{(2n)!}{2n-r \choose r}r!2^r(2n-2r)!-\mathbb{P}\Big((A\cap A_{r+1})\cup \dots \cup (A\cup A_n)\Big) \\ &=&\frac{1}{(2n)!}{2n-r \choose r}r!2^r(2n-2r)!-\sum_{k=1}^{n-r}(-1)^{k-1}{n-r \choose k} \mathbb{P}(A_{1}\cap \dots \cap A_{r+k}) \\ &=& \frac{1}{(2n)!}{2n-r \choose r}r!2^r(2n-2r)!-\frac{1}{(2n)!}\sum_{k=1}^{n-r}(-1)^{k-1}{n-r \choose k} {2n-r-k \choose r+k}2^{r+k}(r+k)!(2n-2r-2k)! \end{eqnarray*}$$ Mutiplying the last expression by ${n \choose r}=\text{# of ways to designate the }r \text{ couples}$ yields the probability we seek: $$\frac{{n \choose r}}{(2n)!}\times \Bigg({2n-r \choose r}r!2^r(2n-2r)!-\sum_{k=1}^{n-r}(-1)^{k-1}{n-r \choose k} {2n-r-k \choose r+k}2^{r+k}(r+k)!(2n-2r-2k)!\Bigg)$$ Showing that this expression matches the one provided in the probability is just a matter of algebraic manipulation.
