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Math Help - Permutations: People in Rooms

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    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?
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  3. #3
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    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?

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