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