Question

(a) Following the methods of Examples $3,4,5$, show that the number of equally likely ways of putting N particles in n boxes,$n>N$, $n^N$isfor Maxwell Boltzmann particles, $C(n,N)$for Fermi-Dirac particles, $C(n-1+N,N)$andfor Bose-Einstein particles.

(b) Show that if n is much larger than N (think, for example, of$n={10}^6,N=10$), then both the Bose-Einstein and the Fermi-Dirac results in part (a) contain products of N numbers, each number approximately equal to n. Thus show that for n N, both the BE and the FD results are approximately equal to$\frac{n^N}{N!}$which is$\frac{1}{N!}$times the MB result.

Short answer

The required values of part (a) are given below.

$\begin{array}{c}M=n^N\\ M=C(n,N)\\ M=C(n-1+N,N)\end{array}$

Part (b) has been proven.

Step-by-step solution

Given Information

The N particles are n boxes.

Definition of uniform sample spaces.

If a given experiment's sample space is known to be uniform, the probability of an event can be calculated using the event sizes and the sample space.

Find the values of part(a).

Use the Maxwell Boltzmann method.

Use n boxes to N particle.

$\begin{array}{c}M=n\times n\times n\times n\mathrm{..}\times n\\ =n^N\end{array}$

Use the Fermi-dirac method. In this method,no more than one ball can be placed in one box and there are N branches among n branches.

$\begin{array}{c}{M}_1=n(n-1)(n-2)\mathrm{..3.2.1}\\ =n!\\ M_2=C(n,N)\end{array}$

The total number is given below.

$\begin{array}{c}{M}_3=M_1{M}_2\\ =n!C(n,N)\end{array}$

Hence $M=C(n,N)$,

Use the Bose Einstein method. In this method, particles are not distinguishable but allow to put 2 balls in the same box. Choose $n$particles among $n+N-1$the equation becomes as follows.

$M_3=C(n-1+N,N)$

Step 3: Prove part(b).

For $n>>N$equation becomes as follows.

$\begin{array}{c}{M}_2=C(n,N)\\ =\frac{n!}{(n-N)!N!}\end{array}$

Formula states that $\frac{n!}{(n-N)!}=n(n-1)(n-2)\dots (n-N+1)$

The equation becomes as follows.

$\begin{array}{c}{M}_2=\frac{n(n-1)(n-2)\dots (n-N+1)}{N!}\\ \approx \frac{n\left(n\right)\left(n\right)\dots \left(n\right)}{N!}\\ \approx \frac{n^N}{N!}\end{array}$

The approximate number of outcomes are given below.

$C(n-1+N,N)=\frac{(n-1+N)!}{N!(n-1)!}$

Formula states that $C(-n,N)={(-1)}^{n}C(n+N-1,N$

Take mod of above equation. The equation becomes as follows.

$\begin{array}{c}C(n+N-1,N)=\frac{n(n+1)(n+2)\dots }{N!}\\ \approx \frac{n*n*n*n\dots }{N!}\\ \approx \frac{n^N}{N!}\end{array}$

Hence, the required values of part (a) are given below.

$\begin{array}{c}M=n^N\\ M=C(n,N)\\ M=C(n-1+N,N)\end{array}$

Part (b) has been proven.