The exact probabilities of equation \((9-9)\) rest on the claim that the number
of ways of adding \(N\) distinct nonnegative integers to give a total of \(M\) is
\((M+N-1) !\) \([M !(N-1) !]\). One way to prove it involves the following trick.
It represents two ways that \(N\) distinct integers can add to \(M-9\) and 5 ,
respectively, in this special case.
$$
\begin{array}{c|cccccccccccccc}
\hline \text { 1 } & \text { X } & \text { X } & \text { X } & \text { I } &
\text { I } & \text { X } & \text { I } & \text { I } & \text { I } & \text {
I } & \text { X } & \text { I } & \text { I } \\
\hline 2 & \text { I } & \text { X } & \text { X } & \text { I } & \text { I }
& \text { I } & \text { I } & \text { X } & \text { I } & \text { I } & \text
{ I } & \text { X } & \text { X } \\
\hline
\end{array}
$$
The X's represent the total of the integers, \(M\) - each row has \(5 .\) The I's
represent "dividers" between the distinct integers, of which there will of
course be \(N-\) I -each row has 8 . The first row says that \(n_{1}\) is 3 (three
\(X\) 's before the divider between it and \(n_{2}\) ). \(n_{2}\) is 0 (no \(X\) 's
between its left divider with \(n_{1}\) and its right divider with
\(\left.n_{3}\right), n_{3}\) is \(1, n_{4}\) through \(n_{6}\) are \(0, n_{2}\) is
\(1,\) and \(n_{8}\) and \(n_{9}\) are
0\. The second row says that \(n_{2}\) is \(2, n_{6}\) is \(1, n_{9}\) is \(2,\) and
all other \(n\) are 0 . Further rows could account for all possible ways that
the integers can add to \(M\). Argue that. properly applied, the binomial
coefficient (discussed in Appendix J) can be invoked to give the correct total
number of ways for any \(N\) and \(M\).
