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?
Hello, Lukybear!
Plato is absolutely correct . . . There are: . ways.
Your error is a very subtle one.
It took me a moment to ind it.
In how many ways can 4 people be assigned to 4 rooms?
I thought it combines both permutation and combination.
I considered:
A. 4 People sharing 1 room .
B. 3 People sharing 1 room: .
C. 2 People sharing 1 room: .
. . C1. The other 2 people are together: .
. . C2. The 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.
The error lies in case
. . the people are divided into two pairs.
Call the people
How many pairings are possible?
The answer is three: .
. . Note that is not a different pairing.
Make this correction and you will get 256.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I first ran into this snag many years ago.
There are four bridge players.
In how many ways can the players be paired?
I immediately said: .
I was surprised to learn that the answer was three.
(And no explanation was given.)
I finally reasoned it out.
When we select a pair of players to be partners,
. . we have automatically assigned the other two to be partners.
Call the players
With whom can be paired?
can be paired with . . . 3 choices.
And that's it!
. . Get it?