Four married couples (4 men and their 4 wives) are available for mixed doubles tennis. A mixed doubles game uses two teams, each team being one man paired with one woman. What is the maximum number of distinct games possible if no team may consist of a husband paired with his own wife?

Introduction / Context:
We must count distinct mixed-doubles games (two disjoint man–woman pairs) from 4 married couples, with the constraint that a husband and his own wife can never be on the same team. A person may appear in many different games across the count; we are counting all possible pairings of two teams adhering to the restriction.

Given Data / Assumptions:

• Men: M1–M4; Women: W1–W4 with couples (Mi, Wi).
• Game = two disjoint pairs {(man, woman), (man, woman)}.
• No pair may be a married couple (Mi, Wi).
• Games are considered unordered collections of two teams (i.e., swapping the two teams does not produce a new game).

Concept / Approach:
Fix the set of 2 men and the set of 2 women and then count valid pairings between them. Sum over all choices of 2 men and 2 women.

Step-by-Step Solution:
Choose men: C(4,2) = 6. Choose women: C(4,2) = 6.For fixed {Mi, Mj} and {Wk, Wℓ}, there are 2 possible pairings: (Mi–Wk & Mj–Wℓ) or (Mi–Wℓ & Mj–Wk).Count valid pairings by cases:• If {Wk, Wℓ} = {Wi, Wj}: one pairing is forbidden (both married), the cross pairing is valid → 1 valid.• If exactly one of {Wk, Wℓ} is a spouse of a selected man: exactly 1 valid pairing avoids that marriage → 1 valid.• If none of {Wk, Wℓ} is a spouse of the chosen men: both pairings valid → 2 valid.Counting across the 6 women-choices for fixed men yields 7 valid pairings; across 6 men-choices: 6 × 7 = 42 games.

Verification / Alternative check:
The bipartite graph K4,4 minus the 4 marriage edges has exactly 42 matchings of size 2 (unordered), consistent with the above casework.

Why Other Options Are Wrong:
12 and 36 undercount; 48 overcounts by treating invalid pairings as valid; 30 is ad hoc.

Common Pitfalls:
Confusing games with single pairs or double-counting by ordering the two teams.

Final Answer:
42