Probability --please help --which approach is correct

The question is quite ambiguous.

There are three people Sam, Rick and Husam. What is the probability that they don't have birthday on a same day (Sunday, Monday, ..., Saturday)?

I think that the question simply asks the probability of an event that not all 3 men have their birthdays on a same day (meaning that all 3 have birthday on Sunday is not OK, but if 2 of them have their birthday on Sunday and the third one on Monday is OK):

\(P=1-\frac{7}{7^3}=\frac{48}{49}\). So in this case neither of approaches is right.

But if the question were asking about the probability that all 3 men have their birthdays on different days (for example Sam-Sunday, Rick-Monday, Husam-Tuesday), then:

First one can have his birthday on any day - 7/7, the probability that the second guy won't have his birthday on the same day will be 6/7 and the probability that the third guy won't have his birthday on these two days will be 5/7, so \(P=\frac{7}{7}*\frac{6}{7}*\frac{5}{7}=\frac{30}{49}\).

Or if you want to solve with combination then: probability=# of favorable outcomes/total # of outcomes --> total # of outcomes is 7^3 as each has 7 options for a birthday. Now, # of favorable outcomes will be: \(C^3_7*3!\), where \(C^3_7\) is ways to choose 3 different days out of 7 and 3! is # of different ways to assign these days to 3 people (the nominator also could be represented as \(P^3_7\)). So, \(P=\frac{C^3_7*3!}{7^3}=\frac{30}{49}\).

So in this case your first approach would be correct and second would not. By the way second approach doesn't give a probability at all (probability must be between 0 and 1 inclusive), and it gives # of selections of 3 different items (days in our case) out of 7.

Hope it's clear.