In a society it is customary for friends of the same sex to hug and for friends met in a party and there were 24 handshakes. Which one among the…

Correct Option (3)

The problem establishes two distinct social customs: handshakes occur exclusively between friends of the opposite sex, and hugs occur between friends of the same sex. We are given that there were 24 handshakes.

• Let 'x' represent the number of males and 'y' represent the number of females in the party.
• Since handshakes occur between males and females, the total number of handshakes is the product of the number of males and females: x × y = 24.
• Considering integer values for x and y that represent reasonable group sizes for a party, the possible pairs (x, y) satisfying x × y = 24 are (4, 6) or (6, 4).
• Hugs occur between friends of the same sex. The number of hugs can be calculated using combinations:
  • Number of hugs among males = _xC₂ (combinations of 2 males chosen from x)
  • Number of hugs among females = _yC₂ (combinations of 2 females chosen from y)
  • Total number of hugs = _xC₂ + _yC₂
• Using the pair (x=6, y=4):
  • ₆C₂ = 6! / (2! × 4!) = (6 × 5) / (2 × 1) = 15
  • ₄C₂ = 4! / (2! × 2!) = (4 × 3) / (2 × 1) = 6
  • Total hugs = 15 + 6 = 21
• If the pair (x=4, y=6) is used, the calculation yields ₄C₂ + ₆C₂ = 6 + 15 = 21, confirming the result.

Thus, the possible number of hugs is 21.

Incorrect Options:

Options 1 (39), 2 (30), and 4 (20) are incorrect. Based on the condition that x × y = 24 handshakes occur between opposite sexes, and considering reasonable party sizes, the only valid combination of males and females (x, y) is (6, 4) or (4, 6). This combination uniquely leads to 21 hugs. Other integer pairs for x × y = 24, such as (3, 8) or (8, 3), would result in ₃C₂ + ₈C₂ = 3 + 28 = 31 hugs, which is not provided as an option. Therefore, 21 is the only number of hugs consistent with the problem's constraints among the given choices.