There are $4$ married couples & No husband can be team up with his wife.
Assuming, $P, Q, R, S$ are the husbands & $A,B,C,D$ are the wives.
So, the married couples are : $(P,A);(Q,B);(R,C);(S,D)$
Now, we'll choose $2$ husbands first from the $4$ husbands i.e. $^4C_2 = 6$
∴ There are $6$ ways we can choose Males for the games.
Now, If we choose $2$ male think $P$ & $Q$, then we'll get $2$ cases:
CASE 1 :
- If $C$ or $D$ is assign with $P$. This can be done in only $2$ ways
- Now, there will be $2$ choices or $2$ female available to be coupled with $Q$.( i.e either $C$ or $D$ and $A$) This can be done in $^2C_1 \text{ ways}$
Total no. of ways of CASE 1: $^4C_2 \times 2 \times ^2C_1 = 6 \times 2 \times 2 = 24$
CASE 2:
- If $B$ is coupled with $P$, can be done in $1$ ways only.
- We've $3$ choices (females ) left with to couple with $Q$ & this can be done in $^3C_1 \hspace{0.1cm} ways$
Total no. of ways of CASE 2: $^4C_2 \times 1 \times ^3C_1 = 6 \times 1 \times 3 = 18$
∴$\color{Green}{\text{ Maximum no. of games that can be played is}}$ = $\text{CASE 1 OR CASE 2}$
$\qquad \qquad \qquad \qquad = \color{teal}{24 + 18 \hspace{0.1cm} ways = 42 \hspace{0.1cm} ways}$