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

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

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

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

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



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

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

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



    3) If all possible arrangements are equally likely,
    find the probability that two of $\displaystyle A, B, C$ 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:
    . . $\displaystyle (1)\;A,B$ are together and $\displaystyle C$ is nonadjacent.
    . . $\displaystyle (2)\;B,C$ are together and $\displaystyle A$ is nonadjacent.
    . . $\displaystyle (3)\;A,C$ are together and $\displaystyle B$ is nonadjacent.


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

    $\displaystyle (a)\;\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\
    \boxed{AB} & \_ & \_ & \_\end{array}$

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


    $\displaystyle (b)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\
    \_ & \boxed{AB} & \_ & \_ \end{array}$

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


    $\displaystyle (c)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\
    \_ & \_ & \boxed{AB} & \_ \end{array}$

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


    $\displaystyle (d)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\
    \_ & \_ & \_ & \boxed{AB} \end{array}$

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


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

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


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

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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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