In a society it is customary for friends of the same sex to hug and for friends of opposite sex to shake hands when they meet. A group of friends met in a party and there were 24 handshakes.

Which one among the following numbers indicates the possible number of hugs?

The correct option is 21.

Let us analyze the problem step-by-step to understand why 21 is a possible number of hugs.

Let the group of friends consist of $m$ males and $f$ females. According to the custom described in the problem:
1. Friends of the opposite sex shake hands when they meet. Therefore, each meeting between a male and a female results in a handshake. The total number of handshakes is the product of the number of males and females:

$m\times f=24$

2. Friends of the same sex hug when they meet. Thus, hugs occur when males meet males, and when females meet females. The number of hugs among males is $\frac{m(m-1)}{2}$, and the number of hugs among females is $\frac{f(f-1)}{2}$. The total number of hugs, $H$, is given by:

$H=\frac{m(m-1)}{2}+\frac{f(f-1)}{2}$

Since $m$ and $f$ must be positive integers, we find the possible pairs of $(m,f)$ such that their product is 24, and calculate the corresponding total hugs $H$ for each case:

Case 1: $m=1$ and $f=24$ (or vice versa)

$H=\frac{1\left(0\right)}{2}+\frac{24\left(23\right)}{2}=0+276=276$

Case 2: $m=2$ and $f=12$ (or vice versa)

$H=\frac{2\left(1\right)}{2}+\frac{12\left(11\right)}{2}=1+66=67$

Case 3: $m=3$ and $f=8$ (or vice versa)

$H=\frac{3\left(2\right)}{2}+\frac{8\left(7\right)}{2}=3+28=31$

Case 4: $m=4$ and $f=6$ (or vice versa)

$H=\frac{4\left(3\right)}{2}+\frac{6\left(5\right)}{2}=6+15=21$

Looking at the options provided, 21 corresponds to Case 4 where there are 4 members of one sex and 6 members of the other sex. Hence, 21 is a possible number of hugs.