Let there be $a$ men and $20-a$ women. The number of pairs in a group of $x$ people is $\frac{(n)(n-1)}{2}$. We have $\frac{a(a-1)+(20-a)(19-a)}{2}=106$. Expanding gives $a^2-a+380-39a+a^2=212 -> 2a^2-40a=-168 -> a^2-20a+84=0$. The two values that satisfy this quadratic are 6 and 14, therefore the answer is $6 \cdot 14 = \boxed{84}$.

Aww.. @above sniped!!

@GeneralCobra19 You don't have to solve that Quadratic equation. As the problem asked to find $a*b$ i.e to find $a(20-a)$ which is nothing but $20a-a^2$. You already got $a^2-20a+84=0$.From there we can get easily get $20a-a^2=84$ which is the answer.

Richie wrote: @GeneralCobra19 You don't have to solve that Quadratic equation. As the problem asked to find $a*b$ i.e to find $a(20-a)$ which is nothing but $20a-a^2$. You already got $a^2-20a+84=0$.From there we can get $20a-a^2=84$ which is the answer. Yeah true. It's still extremely easy to solve though.

Another shortcut way to do this problem is by complementary counting.Since there are $20$ people the total number of handshakes will be ${20\choose2}$. Now this includes the handshakes between men and women as well. We have to subtract that from total. Since there are $a$ men and $b$ women, the number of handshakes between men and women will be $a*b$. So according to the problem ${20\choose2}-(a*b)=106$.This implies $\frac{20*19}{2}-(a*b)=106$ $=>190-(a*b)=106$ $=>a*b=84$