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QUESTION
 A random sample of size n is taken from a population of size N.
   Write down the number of distinct samples when sampling is

    \begin{description}

     \item[(i)]ordered, with replacement

     \item[(ii)]ordered, without replacement

     \item[(iii)]unordered, without replacement

     \item[(iv)]unordered, with replacement

    \end{description}
   (n.b. (\textbf{iv}) is harder, it may help to write down a few
   special cases first).

\bigskip

ANSWER

   \begin{description}

    \item[(i)]There is a choice of $N$ for each therefore $N^n$

    \item[(ii)]$N(N-1)(N-2)\ldots (N-n+1)=NP_n=\frac{N!}{(N_n)!}$

    \item[(iii)]Each choice in this section goes to $n!$ samples in
    (ii), since each unordered sample can be ordered in $n!$ ways.
    Number of samples $=\frac{N!}{(N-n)!n!}=NC_n$

    \item[(iv)]Example, $N=5$ and $n=3$.

     \begin{tabular}{llll}
     &AAA&AAB&ABC\\
     &BBB&AAC&ABD\\
     &CCC&AAD&ABE etc.\\
     total number&5&20&$\left( \begin{array}{c} 5\\3 \end{array} \right)$\\
     \end{tabular}

     Total $=35=\left( \begin{array}{c} 7\\3 \end{array}
     \right) =\left( \begin{array}{c} N+n-1\\n \end{array}
     \right)$
     Problem corresponds first to the classical one of having $N$
     boxes and requiring the placing of $n$ balls with any number in
     each box.

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     \setlength{\unitlength}{.25in}

\begin{center}
\begin{picture}(16,2)
\put(0,0.5){e.g.}
\put(2,0){\framebox(2,1){$\bullet\bullet$}}\put(2.8,1.2){$A$}
\put(5,0){\framebox(2,1){ }} \put(5.8,1.2){$B$}
\put(8,0){\framebox(2,1){$\bullet$}} \put(8.8,1.2){$C$}
\put(11,0){\framebox(2,1){ }} \put(11.8,1.2){$D$}
\put(14,0){\framebox(2,1){ }} \put(14.8,1.2){$E$}
\end{picture}
\end{center}

\smallskip

     This in turn corresponds to the problem of arranging $N+1$
     lines and $n$ circles in a row give that first and last must be
     a line.

\setlength{\unitlength}{.25in}
\begin{center}
\begin{picture}(8,1)
\put(0,-0.5){\line(0,1){1}} \put(1,0){\circle{0.5}}
\put(2,0){\circle{0.5}} \put(3,-0.5){\line(0,1){1}} \put(4,-0.5){\line(0,1){1}}
\put(5,0){\circle{0.5}} \put(6,-0.5){\line(0,1){1}}
\put(7,-0.5){\line(0,1){1}}\put(8,-0.5){\line(0,1){1}}
\end{picture}
\end{center}

\bigskip

     corresponds to $AAC.$ This can be done in $\left( \begin
     {array}{c}N+1-2+n\\n \end{array}\right)$ways (since the first
     and last are fixed.)

    \end{description}

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