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Math Help - probability help

  1. #1
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    probability help

    the question is something like this

    a group of n girls is given names. there are N different names available. The girls are allotted names at random with 2 or more allowed to have same name. show the probability they will have n different names is (N-1)(N-2)....(N-n+1)/N^(n-1)

    i think binomial distribution is involved init somewhere.

    please hellllpppppp
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  2. #2
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    Lexington, MA (USA)
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    Hello, alex_2008!

    A group of n girls is given names.
    There are N different names available.
    The girls are allotted names at random with 2 or more allowed to have same name.

    Show the probability they will have n different names is: . \frac{(N-1)(N-2) \cdots (N-n+1)}{N^{n-1}}
    Let's "walk" through this . . .


    . . . . \begin{array}{ccccccc}\text{The 1st person can have any of the }N\text{ na{m}es:} & \qquad\quad Prob &=& \quad\dfrac{N}{N}\quad&=&\quad 1 \\ \\[-3mm] \end{array}

    \begin{array}{ccccc}\text{The 2nd person can have any of the other }N-1\text{ na{m}es:} & \quad\;\; Prob &=& \dfrac{N-1}{N} \\ \\[-3mm] \end{array}

    \begin{array}{cccccc}\text{The 3rd person can have any of the other }N-2\text{ na{m}es:} & \quad\;\; Prob &=& \dfrac{N-2}{N} \\ \\[-3mm] \end{array}

    \begin{array}{ccccc} \vdots & & \vdots \\<br />
\text{The }n^{th}\text{ person can have any of the other }N-n+1\text{ na{m}es:} & Prob &=& \dfrac{N-n+1}{N} \end{array}


    The probability is: . 1\cdot\underbrace{\frac{N-1}{N}\cdot\frac{N-2}{N} \hdots\frac{N-n+1}{N}}_{n-1\text{ fractions}}


    . . . . . . . . . . . . = \;\frac{(N-1)(N-2)\cdots(N - n+1)}{N^{n-1}}

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  3. #3
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    thanks alot soroban
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