Understanding the Pairing Problem

The problem asks us to form 4 pairs, with each pair consisting of one woman and one man, from a group of 4 women (P, Q, R, S) and 5 men (V, W, X, Y, Z). We also have two important constraints to consider:

• Woman P is not to be paired with Man Z.
• Man Y must necessarily be paired with some woman.

Since there are 4 women and we need to form 4 pairs each with one woman and one man, all 4 women will be part of the pairs. From the 5 men, we must select exactly 4 men to form these pairs.

Selecting the Men for Pairing

The second constraint states that Man Y must necessarily be paired. This means Y is always one of the 4 men selected for pairing. Since Y is already selected, we need to choose 3 more men from the remaining 4 men {V, W, X, Z}.

The number of ways to choose 3 men from these 4 men is given by the combination formula:

$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times 1} = 4$ ways.

These 4 selections of 4 men (including Y) can be categorized into two main cases based on whether Z is included or excluded:

• Case 1: Z is NOT among the 4 selected men.
• Case 2: Z IS among the 4 selected men.

Case 1: Z is Not Selected for Pairing

In this scenario, Z is not part of the group of 4 men chosen for pairing. Since Y is already selected, the remaining 3 men must be chosen from {V, W, X}.

Number of ways to choose 3 men from {V, W, X} = $\binom{3}{3} = 1$ way.

The selected group of 4 men is {V, W, X, Y}.

Because Z is not among the selected men, the constraint "P is not to be paired with Z" is automatically satisfied as Z is not available to be paired with anyone.

Now, we need to form pairs between the 4 women (P, Q, R, S) and these 4 selected men (V, W, X, Y). This is a simple permutation problem of pairing 4 distinct items with another 4 distinct items. The number of ways to arrange 4 women with 4 men is 4! (4 factorial).

$4! = 4 \times 3 \times 2 \times 1 = 24$ ways.

So, there are 24 ways when Z is not selected.

Case 2: Z is Selected for Pairing

In this scenario, Z is one of the 4 selected men. Since Y is also necessarily selected, the remaining 2 men must be chosen from {V, W, X}.

Number of ways to choose 2 men from {V, W, X} = $\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1) \times 1} = 3$ ways.

The possible groups of 4 men (including Y and Z) are:

1. {Y, Z, V, W}
2. {Y, Z, V, X}
3. {Y, Z, W, X}

For each of these 3 sets of men, we need to pair them with the 4 women (P, Q, R, S) while ensuring that P is not paired with Z.

Let's consider one such set of men, for example, {Y, Z, V, W}.

• Total ways to pair 4 women with these 4 men:

  Without any restrictions, the 4 women can be paired with the 4 men in 4! ways.

  $4! = 24$ ways.

• Ways where P IS paired with Z (forbidden pairings):

  If P is paired with Z, then the remaining 3 women (Q, R, S) must be paired with the remaining 3 men (Y, V, W). The number of ways to do this is 3! (3 factorial).

  $3! = 3 \times 2 \times 1 = 6$ ways.

• Ways where P is NOT paired with Z (valid pairings):

  To find the valid pairings for this set of men, we subtract the forbidden pairings from the total pairings:

  Valid ways = Total pairings - (P paired with Z)

  Valid ways = $24 - 6 = 18$ ways.

Since there are 3 such sets of men where Z is selected, the total number of ways for Case 2 is:

Total ways for Case 2 = (Number of sets with Z) $\times$ (Valid ways per set)

Total ways for Case 2 = $3 \times 18 = 54$ ways.

Total Number of Ways to Form Pairs

To find the grand total number of ways to form 4 pairs, we sum the ways from Case 1 and Case 2:

Total Ways = Ways (Z not selected) + Ways (Z selected)

Total Ways = $24 + 54 = 78$ ways.

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