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?


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