Birthday probability: Permutation or Combination? 
In a class of $30$ people, what is the probability that at least $2$ people share the same birthday?

I would think that I should use $^{365}C_{30}$ for the numerator since order shouldn't matter in this case (?) but I have come across $2$ resources that use permutations instead and I am curious as to why.
The resources mentioned above, if anyone needs to refer:
https://www.youtube.com/watch?v=9G0w61pZPig&feature=youtu.be
https://medium.com/i-math/the-birthday-problem-307f31a9ac6f
 A: With permutation, order matters, and with combination, it doesn't.
Imagine you have $30$ people. How many ways are there to assign them unique birthdays?
The question you should ask yourself is: if we assigned Alice $\to$ Jan $10$ and Bob $\to$ Jun $5$, would we consider this a different assignment than if Alice $\to$ Jun $5$ and Bob $\to$ Jan $10$ (for example)? We would, and so order matters. Hence we use permutation.
Another way to see this is to write out your assignments as a $30$-tuple (of days, $1 \ldots 365$), where each spot in the tuple is a specific person. Then you see that $(x, y, \ldots)$ is a different assignment than $(y, x, \ldots)$.
A: When encountering a  problem that says at least count the compliment.
$P(AtLeast2)=1-P(DifferentB-day)$
The probability of any event is counting the number of desired outcomes and divide by the total possibilities. There is $365*364*363...336$ ways for everyone to have a different B-day. And $365^{30}$ total number of B-days
Thus $P(DifferentB-day)$ is $\frac{365*364*363...336}{365^{30}}$
Take the compliment of this and you get your answer
A: The answer is $\approx70,63\%$. Look here and here
Assume we have $n$ people and we ask about the probability that two of the people have their birthdays in the same day. 
If we ask each person when this person was born, then by the time we're done we would have two possible results. 
The first result would be that they all have different birthdays. The second result would be that at least two persons (or at max all $n$ persons) have equal birthdays. 
Let the probability for the first result be $p_1$ and for the second result be $p_2$. Because there are no other results to expect, we have $p_1+p_2=100\%$. 
It is easier to calculate $p_1$, the probability that all birthdays are different from each other: In total we have $365$ days. The first person could have his/her birthday on any of the $365$ days, the second person on any of the remaining $364$ days, ..., the $n$th person on one of the $365-n$ days. Before asking the first person about her/his birthday, we can calculate the probability $p_1^{\text{person 1}}$ that the birthday of the first person is different to all others, this is $100\%$ sure, because we don't know any other birthdays yet. Before asking the second person about her/his birthday, we can calculate the probability for the second person having a different birthday than the others (than the first person). This probability is $p_1^{\text{person 2}}=1-1/365=364/365$, because all days have the same probability $1/365$ to be the birthday of the second person except for one day, except for the day, when person 1 has his birthday, if we want to know the probability of different birthdays for all persons.This goes on and on and on for all $n$ persons and we get finally
$$p_1=\underbrace{1\cdot 364/365 \cdot 363/365 \cdot \cdots \cdot (365-(n-1))/365}_{n \text{ times}}$$
which translates to
$$p_1 = (1\cdot 364\cdot 363 \cdot \cdots \cdot (365-n+1))/365^n = {}_{365}P_n/365^n$$
And the answer to the question is:
$$p_2= 1-{}_{365}P_n/365^n$$
For $n=30$ we get $p_2\approx70,63\%$
