The exact probabilities of equation (9-9) rest on the claim that the number of ways of adding$\mathit{N}$distinct non-negative integer to give a total of$\mathit{M}$ is $\frac{\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{N}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathbf{[}\mathbf{M}\mathbf{!}\mathbf{(}\mathbf{N}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}\mathbf{]}}$ . One way to prove it involves the following trick. It represents two ways that $\mathit{N}$distinct integers can add to$\mathit{M}\mathbf{-}\mathbf{9}$ and$\mathbf{5}$, respectively. In this special case.

The X's represent the total of the integers, $\mathit{M}$-each row has $\mathbf{5}$. The $\mathbf{1}\mathbf{\text{'}}\mathit{s}$represent "dividers" between the distinct integers of which there will of course be $\mathit{N}\mathbf{-}\mathit{I}$each row has$\mathbf{8}$ . The first row says that${\mathit{n}}_{\mathbf{1}}$ is$\mathbf{3}$ (three $\mathit{X}\mathbf{\text{'}}\mathit{s}$before the divider between it and${\mathit{n}}_{\mathbf{2}}$ ), ${\mathit{n}}_{\mathbf{2}}$is$\mathbf{0}$ (no$\mathit{X}\mathbf{\text{'}}\mathit{s}$ between its left divider with${\mathit{n}}_{\mathbf{1}}$ and its right divider with${\mathit{n}}_{\mathbf{3}}$ ),${\mathit{n}}_{\mathbf{3}}$ ) is$\mathbf{1}$ . ${\mathit{n}}_{\mathbf{4}}$through${\mathit{n}}_{\mathbf{6}}$ are$0$ , ${\mathit{n}}_{\mathbf{7}}$is$\mathbf{1}$ , and ${\mathit{n}}_{\mathbf{8}}$and ${\mathit{n}}_{\mathbf{9}}$are$\mathbf{0}$ . The second row says that ${\mathit{n}}_{\mathbf{2}}$is $\mathbf{2}$. ${\mathit{n}}_{\mathbf{6}}$is $\mathbf{1}$, ${\mathit{n}}_{\mathbf{9}}$is$\mathbf{2}$ , and all other$\mathit{n}$ are$\mathbf{0}$ . Further rows could account for all possible ways that the integers can add to$\mathit{M}$ . Argue that properly applied, the binomial coefficient (discussed in Appendix$\mathit{J}$ ) can be invoked to give the correct total number of ways for any$\mathit{N}$ and$\mathit{M}$ .

Step by step solution

01

Given data

The binomial coefficient can be utilized to calculate the number of ways for which$N$ nonnegative integers can add up to$M$ .

02

concept of theK>>k  Approximation. 

$\frac{\mathbf{K}\mathbf{!}}{\mathbf{(}\mathbf{K}\mathbf{-}\mathbf{k}\mathbf{)}\mathbf{!}}\mathbf{\approx }{\mathit{K}}^{\mathbf{k}}$

03

Determine the K>>k Approximation. 

$\begin{array}{c}P=\frac{M!(N-1)!\left((M-n)+(N-1)-1\right)!}{(M-n)!(N-2)!(M+N-1)!}\\ =\frac{M!}{(M-n)!}\times \frac{(N-1)!}{(N-2)!}\times \frac{(M+N-1-(n+1\left)\right)!}{(M+N-1)!}\\ =\frac{M!}{(M-n)!}\times \frac{(N-1)(N-2)!}{(N-2)!}\times \frac{(M+N-1-(n+1\left)\right)!}{(M+N-1)!}\\ =(N-1)\times \frac{M!}{(M-n)!}\times \frac{(M+N-1-(n+1\left)\right)!}{(M+N-1)!}\end{array}$

Solving further, we get

$\begin{array}{c}P=(N-1)\times M^n\times \frac{1}{{(M+N-1)}^{n+1}}\\ =\frac{(N-1)M^n}{{(M+N-1)}^{n+1}}\\ =\frac{N{M}^n}{{(M+N)}^{n+1}}\\ =\frac{N{M}^n}{(M+N){(M+N)}^n}\end{array}$

Simplifying further we get,

$\begin{array}{c}P=\frac{N}{M+N}\times \frac{M^n}{{(M+N)}^n}\\ =\frac{N}{M+N}\times {\frac{M}{M+N}}^n\\ =\frac{N}{M+N}\times {\frac{M+N}{M}}^{-n}\\ =\frac{N}{M+N}\times 1+{\frac{N}{M}}^n\\ =\frac{N}{M+N}{\left[e^{\ln (1+\frac{N}{M})}\right]}^{-n}\\ =\frac{N}{M+N}{e}^{-n\ln (1+\frac{N}{M})}\end{array}$

Thus, the value of P is $\frac{N}{M+N}{e}^{-n\ln (1+\frac{N}{M})}$

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