In how many ways can 4 people be accommodated if there are 4 rooms available?

I though it combines both permutation and combination

I considered

A. 4 People sharing 1 room

4P1

B. 3 People sharing 1 room

4C3 x 4P2

C. 2 People sharing 1 room:

4C2 x

C.1. other 2 people are together:

4P2

C.2. Other 2 people are separate

4P3

D. All 4 people in different rooms:

4P4

Summing, all possibilities together i got 292, which is not the correct answer. Can someone explain the flaw in my arrangment, or perhaps what i missed?