There are 4 women P, Q, R, S, and 5 men V, W, X, Y, Z in a group. We are required to form pairs each consisting of one woman and one man. P is not to be paired with Z, and Y must necessarily be paired with someone. In how many ways can 4 such pairs be formed?
Options:
(A) 74
(B) 76
(C) 78
(D) 80
Official Answer Key:
C
Step-by-Step Solution & Explanation
• 4 pairs use all 4 women and 4 of the 5 men (one man is left unpaired). ‘Y must necessarily be paired’ means Y cannot be the one left out, so the unpaired man is chosen from {V, W, X, Z}.
• Ignoring the P–Z restriction: choose the left-out man from {V,W,X,Z} (4 ways), then arrange the remaining 4 men (which include Y) with the 4 women in 4! = 24 ways. Total = 4×24 = 96.
• Now subtract the arrangements where P is paired with Z (a forbidden pairing): fix P–Z. The remaining 3 women {Q,R,S} must pair with 3 of the remaining 4 men {V,W,X,Y}; since Y must be included, the left-out man now comes from {V,W,X} (3 ways), and the remaining 3 men (including Y) arrange with 3 women in 3! = 6 ways. Total invalid = 3×6 = 18.
• Valid pairings = 96 – 18 = 78.
• Final Answer: (C) 78
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