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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

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