In how many ways can 4 people be accommodated if there are 4 rooms available?
I though it combines both permutation and combination
A. 4 People sharing 1 room
B. 3 People sharing 1 room
4C3 x 4P2
C. 2 People sharing 1 room:
C.1. other 2 people are together:
C.2. Other 2 people are separate
D. All 4 people in different rooms:
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?