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Math Help - probability

  1. #1
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    probability

    Five friends, A,B,C,D and D, stand in a line for a photograph.

    1) How many different possible arrangements are there if A , B and C stand next to each other?

    2) How many different possible arrangements are there if none of A, B and C stand next to each other?

    3) if all possible arrangements are equally likely, find the probability that two of A, B and C are next to each other, but the third is not next to either of the other two.
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  2. #2
    Eater of Worlds
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    You have A,B,C,D,E.

    Tie A,B,C together, then you have 3 items you can arrange in 6 ways.

    But A,B,C can be arranged 6 ways among themselves.

    6*6=36
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  3. #3
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    probability

    thanks a lot

    and what about the remaining questions?

    Thanks
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  4. #4
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    Hello, badi6!

    Five friends, \{A,B,C,D, E\}, stand in a line for a photograph.

    1) How many different possible arrangements are there
    if A, B, C stand next to each other?
    galactus is absolutely correct . . .

    Tie A,B,C together and there are three "people" to arrange: . \boxed{ABC}, D, E
    . . There are: . 3! ways.

    But A,B,C can be ordered in 3! ways.

    . . Therefore, there are: . 3! \times 3! \:=\:\boxed{36} ways.



    2) How many different possible arrangements are there
    if none of A, B, C stand next to each other?
    Place D and E in positions #2 and #4: . \_\:D\:\_\:E\:\_
    . . Then there are 3! ways to place A,B,C.

    But D and E can be placed like this: . \_\:E\:\_\:D\:\_
    . . And there are 3! more ways to place A,B,C.

    Therefore, there are: . 3! + 3! \:=\:12 ways.



    3) If all possible arrangements are equally likely,
    find the probability that two of A, B, C are next to each other,
    but the third is not next to either of the other two.
    I have a laundry-list approach to this one . . .

    There are three cases:
    . . (1)\;A,B are together and C is nonadjacent.
    . . (2)\;B,C are together and A is nonadjacent.
    . . (3)\;A,C are together and B is nonadjacent.


    Case (1): A,B are together and C is nonadjacent.
    Then there are four scenarios to consider . . .

    (a)\;\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\<br />
\boxed{AB} & \_ & \_ & \_\end{array}

    A,B can be reversed (two choices).
    C can be in 3 or 4 (two choices).
    Then D and E can be placed (two choices).
    . . There are: .  2 \times 2 \times 2 \:=\:8 ways.


    (b)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\<br />
\_ & \boxed{AB} & \_ & \_ \end{array}

    A,B can be reversed (two choices).
    C must be in 4 (one choice).
    Then D and E can be placed (2 choices).
    . . There are: . 2 \times 1 \times 2 \:=\:4 ways.


    (c)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\<br />
\_ & \_ & \boxed{AB} & \_ \end{array}

    A,B can be reversed (two choices).
    C must be in 1 (one choice).
    Then D and E can be placed (two choices).
    . . There are: . 2 \times 1 \times2 \:=\:4 ways.


    (d)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\<br />
\_ & \_ & \_ & \boxed{AB} \end{array}

    A,B can be reversed (two choices).
    C can be in 1 or 2 (two choices).
    Then D and E can be placed (2 choices).
    . . There are: . 2 \times 2 \times 2 \:=\:8 ways.


    Hence, there are: . 8 + 4 + 4 + 8 \:=\:24 ways
    . . to have A, B together and C nonadjacent.

    Since there are 3 cases, there are: . 3 \times 24 \:=\:72 ways
    . . to have two of A,B,C together and the third nonadjacent.


    Since there are 5!=120 possible arrangements,
    . . the probability is: . \frac{72}{12} \;=\;\boxed{\frac{3}{5}}

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  5. #5
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    probability

    many thanks for your help!

    Plz could you also hep as much as you could to tackle the quetion number 5!

    http://www.robinson.homechoice.co.uk...on%20Paper.pdf

    Thanks
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