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There are 12 months. First person can have month of birth any one out of 12 months.
Second person, to have a different month of birth, can have any of remaining 
 11 months out of 12 months.
Third person, to have a month of birth different from above two months, can
have any of remaining 10 months out of 12 months.
Fourth person, to have a month of birth different from above three months, can
have any of remaining 9 months out of 12 months.

Hence the probability that four persons have different month of birth s
(12/12) x (11/12) x (10/12)x(9/12)
0 0 0
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