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  1. #1
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    counting

    A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead. How many such garlands(distinguishable) are possible?plz discuss.
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  2. #2
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    Quote Originally Posted by Navesh
    A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead. How many such garlands(distinguishable) are possible?plz discuss.
    Hello,

    you've got exactly 5 places where you can put a bead.

    For the first red bead you can choice out of 5 places, for the second red bead there are only 4 places left. Similar to the two yellow beads and the blue on:

    That means you've got 5*4*3*2*1 = 120 possibilities. BUT : Because you can't distinguish between the first or the second red bead, and you you can't distinguish between the first or the second yellow bead, the amount of possibilities is 4 times to great. So finally:

    \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{2\cdot 2}=30

    Greetings

    EB
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  3. #3
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    Quote Originally Posted by earboth
    Hello,

    you've got exactly 5 places where you can put a bead.

    For the first red bead you can choice out of 5 places, for the second red bead there are only 4 places left. Similar to the two yellow beads and the blue on:

    That means you've got 5*4*3*2*1 = 120 possibilities. BUT : Because you can't distinguish between the first or the second red bead, and you you can't distinguish between the first or the second yellow bead, the amount of possibilities is 4 times to great. So finally:

    \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{2\cdot 2}=30

    Greetings

    EB
    It seems that the problem has been solved as if the beads are being arranged along a row. Won't the circular geometry affect the result?
    Last edited by Navesh; June 25th 2006 at 07:46 PM.
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  4. #4
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    Hello, Navesh!

    You're right . . . EB didn't consider the circular arrangement.


    A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead.
    How many such garlands (distinguishable) are possible?

    I've approached this a number of ways,
    . . and I've finally come up with the right answer (4).

    For the moment, imagine that the beads are of five different colors.

    Since they are arranged in a circle, the first bead can be placed anywhere.
    Then the four remaining beads can arranged in 4! = 24 ways.

    But we do not have five different colors; there is duplication.
    The two red beads can be switched without creating a new arrangement.
    The two yellow beads can be switched without creating a new arrangment.
    Hence, there are: \frac{24}{2\cdot2} = 6 ways.


    Now, the garland can be "flipped".
    That is:
    . . . . . + - - * - - - - - - -
    . . . . . * - - - - - * - - - -
    . . . . . + - - - - - - - - - -
    . . . . . + - - - - - - - - - -
    . . . . . + - * - * - - - - - -





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  5. #5
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    Hello, Navesh!

    You're right . . . EB didn't consider the circular arrangement.


    A garland is to be made of 2 red beads, 2 yellow beads and 1 blue bead.
    How many such garlands (distinguishable) are possible?

    I've approached this a number of ways,
    . . and I've finally come up with the right answer (4).

    For the moment, imagine that the beads are of five different colors.

    Since they are arranged in a circle, the first bead can be placed anywhere.
    Then the four remaining beads can arranged in 4! = 24 ways.

    But we do not have five different colors; there is duplication.
    The two red beads can be switched without creating a new arrangement.
    The two yellow beads can be switched without creating a new arrangment.
    Hence, there are: \frac{24}{2\cdot2} = 6 ways.


    Now, the garland can be "flipped over"
    . . say, left to right over a vertical axis of symmetry.

    . . . . . . . . . . .R . . . . . . . . . . . . . . . . . . . . R
    . . . . . . . B . . . . . R. . . . . . . . . . . . . . .R . . . . . B
    That is:. . . . . . . . . . . is the same as:

    . . . . . . . . .Y . . Y . . . . . . . . . . . . . . . . . Y . . Y

    So I assumed that I would divide the remaining 6 arrangements by 2.
    . . But the answer not 6 \div 2 = 3, but 4.
    It seems that the six arrangements are not two sets of mirror-images.


    I found this quite difficult to justify . . . then I found a baby-talk approach.


    The two Reds and two Yellows can be placed in two arrangements:

    . . . . R . . R . . . . . . . R . . Y
    . . . . . . . . . . . and . . . . . . . . disregarding rotations and reflections.
    . . . . Y . . Y . . . . . . . Y . . R


    . . . . R .*. R . . . In the first arrangement, the Blue can be placed
    . . . . . . . .* . . . .in three positions: between the Reds, between
    . . . . Y .*. Y . . . . the Yellows, or between a Red and a Yellow.


    . . . . R .*. Y . . In the second arrangement, the Blue can be placed in
    . . . . . . . . . . . . exactly one position: between a Red and a Yellow.
    . . . . Y . . R . . . . . . . (The rest are rotations and reflections.)


    And there are the four possible arrangements.

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