Lets assume number of children are X
number of ladies are Y
number of gents are Z
Now
X+Y+Z = 20 ;
1/2 x + 2 Y + 3 Z = 20 ;
and X,Y,Z are integers and greater then or equal to zero.
Now we can deduce -
X is even
y < 10 and Z < 6
Now try trial and error after removing one variable from able two equations (i.e. 3y+5z = 20) which gives as y=0 & z=4 and y=5 & z=1. Second solution fits in the above equations so
number of children X = 14
number of ladies Y = 5
number of gents Z = 1