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Math Help - combinations

  1. #1
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    combinations

    The letters P.O.S.S.E.S.S.E.S are shuffled and four selected and placed in a line.
    How many possible selections are there of four letters.

    I did it the long way by coming up with each case. For example,
    case 1 - no 's'
    P.O.E.E
    case 2 - one 's'
    S.P.O.E; S.P.E.E; S.O.E.E
    case 3 - two 's'
    S.S.P.E; S.S.E.E; S.S.O.E; S.S.P.O
    case 4 - three 's'
    S.S.S.E; S.S.S.P; S.S.S.O
    case 5 - four 's'
    S.S.S.S
    This adds up to be 12.
    Is there a better way to do this??

    Thanks.
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  2. #2
    MHF Contributor
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    Re :

    Quote Originally Posted by woollybull View Post
    The letters P.O.S.S.E.S.S.E.S are shuffled and four selected and placed in a line.
    How many possible selections are there of four letters.

    I did it the long way by coming up with each case. For example,
    case 1 - no 's'
    P.O.E.E
    case 2 - one 's'
    S.P.O.E; S.P.E.E; S.O.E.E
    case 3 - two 's'
    S.S.P.E; S.S.E.E; S.S.O.E; S.S.P.O
    case 4 - three 's'
    S.S.S.E; S.S.S.P; S.S.S.O
    case 5 - four 's'
    S.S.S.S
    This adds up to be 12.
    Is there a better way to do this??

    Thanks.
    i misintepreted , plato has corrected ..
    Last edited by mathaddict; February 7th 2009 at 05:15 AM.
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  3. #3
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    Quote Originally Posted by woollybull View Post
    The letters P.O.S.S.E.S.S.E.S are shuffled and four selected and placed in a line.
    How many possible selections are there of four letters.

    I did it the long way by coming up with each case. For example,
    case 1 - no 's'
    P.O.E.E
    case 2 - one 's'
    S.P.O.E; S.P.E.E; S.O.E.E
    case 3 - two 's'
    S.S.P.E; S.S.E.E; S.S.O.E; S.S.P.O
    case 4 - three 's'
    S.S.S.E; S.S.S.P; S.S.S.O
    case 5 - four 's'
    S.S.S.S
    This adds up to be 12.
    Is there a better way to do this??
    It may not be better but it is quicker: \frac{4!}{2!} = 12

    That answer is just for no sís.Other cases are: Exactly one s: <SPOE>, <SPEE>, <SOEE>
    Exactly two sís: <SSOE>, <SSEE>, <SSOP>, <SSEP>
    Exactly three sí: ?
    Exactly four sí: ?
    Last edited by Plato; February 7th 2009 at 05:44 AM.
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  4. #4
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    I don't have easy internet access so I'm trying to interpret what I missed.
    Is there a better way?
    Also I am after combinations - not permutations (arrangements).
    Thanks.
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  5. #5
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    Hello, woollybull!

    The letters of POSSESSES are shuffled.
    Four are selected and placed in a line.
    How many possible selections are there of four letters?
    Since they are "placed in a line", I assume that their order is important.
    . . That is, we are spelling four-letter "words."

    I see no way avoid a brute-force listing . . .


    \begin{array}{ccccc}<br />
\text{No S's} \\ \hline \\[-4mm]<br />
\text{Letters: }P, O, E, E & &\frac{4!}{2!} &=& 12\text{ words}\end{array}

    \begin{array}{ccccc}<br />
\text{One S} \\ \hline \\[-4mm]<br />
\text{Letters: }P, O, E, S & & 4! &=& 24\text{ words} \\<br />
\text{Letters: }P, E, E, S & & \frac{4!}{2!} &=& 12\text{ words} \\ \\[-4mm]<br />
\text{Letters: }O, E, E, S & & \frac{4!}{2!}&=& 12\text{ words} \end{array}

    \begin{array}{ccccc}<br />
\text{Two S's} \\ \hline \\[-4mm]<br />
\text{Letters: P, O, S, S} & & \frac{4!}{2!} &=& 12\text{ words} \\ \\[-4mm]<br />
\text{Letters: }P, E, S, S & & \frac{4!}{2!}&=& 12\text{ words} \\ \\[-4mm]<br />
\text{Letters: }O, E, S, S & & \frac{4!}{2!} &=& 12\text{ words} \\ \\[-4mm]<br />
\text{Letters: }E, E, S, S & & \frac{4!}{2!\,2!} &=& 6\text{ words} \end{array}

    \begin{array}{ccccc}<br />
\text{Three S's} \\ \hline \\[-4mm]<br />
\text{Letters: }P, S, S, S & & \frac{4!}{3!} &=& 4\text{ words} \\ \\[-4mm]<br />
\text{Letters: }O, S, S, S & & \frac{4!}{3!} &=& 4\text{ words} \\ \\[-4mm]<br />
\text{Letters: }E, S. S. S & & \frac{4!}{3!} &=& 4\text{ words} \end{array}

    \begin{array}{ccccc}<br />
\text{Four S's} \\ \hline \\[-4mm]<br />
\text{Letters: }S, S, S, S & & \frac{4!}{4!} & =& 1\text{ word} \end{array}


    Now add them up . . .

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