Birthday Problem: Hypothetically Not Using the Complement My Professor said that the probability that at least 2 of $k$ people have the same birthday if we didn’t use the complement is the sum of the probabilities that exactly 2 have the same birthday, plus the probability 3 have the same birthday, and so on…
However, I wanted to verify this and believe I thought of a case where this is not true: the case where exactly 3 and exactly 4 have the same birthday. I.e. the event that exactly 2 and exactly 3 people have the same birthday are not disjoint. therefore, we need a different method to calculate the probability not using the complement. Is this valid?
 A: Comment continued: Suppose there are 365 equally likely birthdates in a year (ignoring instances of Feb. 29) and that 23 people have randomly chosen birthdays.
n =23;  m = 10^6
set.seed(2022)
x = replicate(m, 
              n-length(unique(sample(1:365, n, rep=T))))
mean(x == 0)
[1] 0.492107     # aprx P(No Match)
2*sd(x == 0)/1000
[1] 0.0009998759 # aprx 95% margin of sim error

So the simulated probability of no matches is $0.4921 \pm 0.0010.$ The exact probability is $0.4927.$
[An advantage of simulation is that one can easily
get the probability of no matches based on data
for the relevant country of the actual probabilities
of different birthdates (often slightly higher in summer than in winter).]
prod(1-(0:22)/365)
[1] c028    # exact P(No Match)

Here is a histogram of the simulated distribution of the number of matches among
23 people.

cutp = (-1:max(x))+.5
hist(x, prob=T, br=cutp, col="skyblue2")

Addendum: The alternative simulation below
may provide an additional perspective:
n =23;  m = 10^6
set.seed(2022)
x = replicate(m, 
             sum(duplicated(sample(1:365, n, rep=T))))
mean(x == 0)
[1] 0.492107

A: If you have a random variable $X$ (e.g., the maximum number of people in the sample that share a birthday) that can only take on values $0,1,2,...,k$ then it is certainly true that:
$$\{ X \geqslant 2 \} = \bigcup_{x=2}^k \{ X = x \}.$$
Consequently, you then have:
$$\mathbb{P}(X \geqslant 2) = \sum_{x=2}^k \mathbb{P}(X = x).$$
Presumaby this is what your professor is talking about in this case.  Your misgivings seem to arise from ambiguity about the event whose probability you are computing.  If you can clarify that with your professor then this might become clearer.
