A certain planet has n days in one year. What is the probability that among $k$ people on that planet there are (at least) two who share their birthday?
My answer to this practice question is: There are $N^k$ probabilities/cases in total. We now have to count the favorable/conditional cases. There are ${n}\choose{k}$ ways to select no two people having birthday on the same day. The probability is then $p=1-\frac{{n}\choose{k}}{N^k}$.
But I'm sure the problem is more complicated than that...
While $k>n$, it is obviously a $100$ percent chance, so we shall not consider a case. So, as the answer says, the number of undesired outcomes is $({n\atop k})$, because we are choosing $k$ different days from $n$ to ensure that there is no overlapping of birthdays. Note the importance of the world different, as it means that there are $k$, and only $k$ days to have the birthdays, or else there would be overlaps.
Then I presume the rest would be easily understandable, because it is just finding the fraction of possible ways over the total ways.