Why isn't adding the ways to achieve every mutually exclusive outcome giving me the denominator in the birthday problem for four people?
$$\binom{4}{2} \cdot 365 \cdot 364 +\binom{4}{3} \cdot 365 \cdot 364+365 \cdot 364 \cdot 363 \cdot 362+365-365^4 \neq 0$$ :
Exactly Two share a Birthday + Exactly Three share a Birthday + None Share a Birthday + All Share a Birthday - Denominator of Birthday problem for 4 People Should Equal Zero.
I expect this to work because the sum of binomial coefficients in pascal's should equal the $2^{row}.$
The corrected calculation is as follows:
$\underbrace{\binom{4}{2}\cdot 365\cdot 364\cdot 363}_{\text{aabc}} + \underbrace{\binom{4}{2}\cdot\binom{365}{2}}_{\text{aabb}}+\underbrace{4\cdot 365\cdot 364}_{\text{aaab}}+\underbrace{365\cdot 364\cdot 363\cdot 362}_{\text{abcd}}+\underbrace{365}_{\text{aaaa}} = 365^4$