# Number of ways to split 8 people into groups of 2

In how many ways can we split 8 people who are married into 4 (unlabeled) groups of two (each group consists of one man and one woman) such that no group is a married couple? So a solution is a set $$\{(M_1,W_1),(M_2,W_2),(M_3,W_3),(M_4,W_4)\}$$. Hint: for $$i=1,2,3,4$$, let $$A_i$$ be the set of groupings $$\{(M_1,W_1),(M_2,W_2),(M_3,W_3),(M_4,W_4)\}$$ such that married couple $$i$$ is one of the groups.

• Hint: look up "derangements". – Gerry Myerson Sep 21 '18 at 7:05

Let $$A_i=\{\text{set of groupings such that married couple }i\text{, for i=1,2,3,4 is one of the groups}\}$$ and $$X=\{\text{# of ways to split 4 married couples in groups of 2 with one man and one woman}\}$$. We know that $$|X|=4!$$ because if we first fix the 4 men, then there are 4! ways of seating the 4 women. For each $$A_i$$, there are $$\left(\begin{array}{c} 4 \\ 1 \end{array}\right)$$ ways to fix one married couple and then 3! ways to seat the remaining 3 women. Then the intersection of any two $$A_i$$'s would be $$\left(\begin{array}{c} 4 \\ 2 \end{array}\right)$$ ways to fix two married couples with 2! ways to seat the remaining 2 women. Similarly, for the intersection of three $$A_i$$'s, we get $$\left(\begin{array}{c} 4 \\ 3 \end{array}\right)1!$$ and the intersection of all 4 $$A_i$$'s would be $$\left(\begin{array}{c} 4 \\ 4 \end{array}\right)0!$$. By PIE, we have \begin{align*} |X\backslash(A_1\cup A_2\cup A_3\cup A_4)|=&4!-3!\left(\begin{array}{c} 4 \\ 1 \end{array}\right)+2!\left(\begin{array}{c} 4 \\ 2 \end{array}\right)-1!\left(\begin{array}{c} 4 \\ 3 \end{array}\right)+0!\left(\begin{array}{c} 4 \\ 4 \end{array}\right)\\ =&24-24+12-4+1\\ =&9. \end{align*}