Pairing Calculation: Women and Men Groups

The problem asks for the number of ways to form 4 pairs, each with one woman and one man, from 4 women (P, Q, R, S) and 5 men (V, W, X, Y, Z), subject to specific constraints.

1. Total Possible Pairings

First, calculate the total ways to form 4 pairs without restrictions. We need to select 4 men out of 5 to be paired, and then arrange the pairings between the 4 women and the 4 selected men.

• Ways to choose 4 men from 5: $\binom{5}{4} = 5$
• Ways to pair the 4 women with the 4 chosen men: $4! = 4 \times 3 \times 2 \times 1 = 24$
• Total unrestricted pairings ($N$): $N = \binom{5}{4} \times 4! = 5 \times 24 = 120$

2. Applying Constraints using Inclusion-Exclusion

We need to satisfy two conditions: P is NOT paired with Z, and Y MUST be paired.

Let's use the principle of inclusion-exclusion. We start with the total unrestricted pairings (120) and subtract the pairings that violate *at least one* of the conditions.

Let A be the set of pairings where P is paired with Z.

Let B be the set of pairings where Y is unpaired.

We want to find the number of pairings where P is NOT with Z (complement of A) AND Y IS paired (complement of B). This is given by $N - |A \cup B|$.

Calculate $|A|$: Pairings where P is paired with Z

• If P is paired with Z, we need to form 3 more pairs.
• The remaining women are {Q, R, S} and remaining men are {V, W, X, Y}.
• Choose 3 men from {V, W, X, Y} to be paired: $\binom{4}{3} = 4$ ways.
• Pair the 3 women {Q, R, S} with the 3 chosen men: $3! = 6$ ways.
• $|A| = \binom{4}{3} \times 3! = 4 \times 6 = 24$

Calculate $|B|$: Pairings where Y is unpaired

• If Y is unpaired, the 4 women {P, Q, R, S} must be paired with the remaining 4 men {V, W, X, Z}.
• The number of ways to pair these 4 women with these 4 men is $4! = 24$.
• $|B| = 4! = 24$

Calculate $|A \cap B|$: Pairings where P is paired with Z AND Y is unpaired

• If P is paired with Z and Y is unpaired, we need to form 2 more pairs.
• The remaining women are {Q, R, S} and the remaining men are {V, W, X}.
• Pair the 3 women {Q, R, S} with the 3 men {V, W, X}: $3! = 6$ ways.
• $|A \cap B| = 3! = 6$

Calculate $|A \cup B|$: Pairings violating at least one condition

• Using the inclusion-exclusion principle: $|A \cup B| = |A| + |B| - |A \cap B|$
• $|A \cup B| = 24 + 24 - 6 = 42$

3. Final Calculation

The number of ways satisfying both conditions (P not with Z AND Y paired) is the total ways minus the ways violating at least one condition:

• Ways = $N - |A \cup B|$
• Ways = $120 - 42 = 78$

Therefore, there are 78 ways to form the 4 pairs according to the given conditions.