probability - Solving birthday problem without complement

I'm trying to find the probability of at least 2 people in a room of 4 sharing the same birthday (without using complements).

I began by breaking the problem down into 4 cases:

Let E = the event that at least 2 people share the same birthday in a room of 4.

Our sample size: $365^4$

Case 1: 4 people share the same birthday: 365 ways

Case 2: 3 people share the same birthday, 1 distinct birthday: $365 \cdot 364 \cdot C(4,3)$

Case 3: 2 people share a birthday, another 2 people share some other birthday: $365 \cdot 364 \cdot \frac{C(4,2)}{2}$

Case 4: 2 people share same birthday, 2 distinct birthdays: $365 \cdot 364 \cdot 363 \cdot C(4,2) \cdot 2$

After adding up all the cases and dividing by the sample size to find probability the answer had an over-count. I checked my answer by doing $$P(E) = 1- \frac{365 \cdot 364 \cdot 363 \cdot 362}{365^4}$$

Where did I have an over-count? Thank you!

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Here is an example that works with n = 3 people and at least 2 people share same birthday.

Case 1: 3 people share same birthday: 365

Case 2: 2 Same birthdays, 1 different: $365 \cdot 364 \cdot \binom{3}{2}$

$$P(E) = \frac{365 + (365 \cdot 364 \cdot \binom{3}{2})}{365^3} \equiv 1 - \frac{365 \cdot 364 \cdot 363}{365^3}$$

Those are both equivalent answers because in the complement we're subtracting away the event that all birthdays are distinct.