Suppose all 20 births are equally likely to be in any of the 12 months of the year and all are independent.
Probability that JAN,FEB,MAR,APR have 2 birthdays and MAY,JUN,JUL,AUG have 3 birthdays and other months have 0 months =
=1/$12^{20}$ * 20!/($(2!)^{4}*(3!)^{4}*(0!)^{4}$ $\approx$ 3 * 10$^{-8}$
Now other ways in which we can choose months that contain 2 birthday or 3 birthdays or 0 birthdays=12!/(4!)$^{3}$
So final probability=3 * 10$^{-8}$ * 12!/(4!)$^{3}$=0.001