# Thread: Permutations: People in Rooms

1. ## Permutations: People in Rooms

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?

2. Originally Posted by Lukybear
In how many ways can 4 people be accommodated if there are 4 rooms available?
Without any restrictions the answer is clearly $4^4$!
That is the number of functions from a set of four to a set of four.

But I do not think that what you have in mind.
Is it?

3. Hello, Lukybear!

Plato is absolutely correct . . . There are: . $4^4 = 256$ 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 . $_4P_1$

B. 3 People sharing 1 room: . $(_4C_3) \times (_4P_2)$

C. 2 People sharing 1 room: . $(_4C_2) \times$

. . C1. The other 2 people are together: . $_4P_2$

. . C2. The other 2 people are separate: . $_4P_3$

D. All 4 people in different rooms:: . $_4P_4$

Summing all possibilities together, i got 292
. . which is not the correct answer.

The error lies in case $C1:$
. . the people are divided into two pairs.

Call the people $A,B,C,D.$

How many pairings are possible?

The answer is three: . $\{AB|CD\},\;\{AC|BD\},\;\{AD|BC\}$

. . Note that $\{CD|AB\}$ 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: . $_4C_2 \:=\:\dfrac{4!}{2!\,2!} \:=\:6$

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 $A,B,C,D.$

With whom can $A$ be paired?

$A$ can be paired with $B,C,\text{ or }D$ . . . 3 choices.

And that's it!

. . Get it?