Let r particles be distributed among n cells according to…

Let $r$ particles be distributed among $n$ cells according to Bose–Einstein statistics (see Problem 58).
$1^{\circ}$ Prove that the probability that exactly $k$ particles are in a fixed cell is \[ q_k=\frac{\binom{n+r-k-2}{r-k}}{\binom{n+r-1}{r}}. \] $2^{\circ}$ Show that $q_0\gt q_1\gt q_2\gt\dots$
$3^{\circ}$ Prove that if $n$ and $r$ grow without bound while the average number of particles per cell $\frac rn$ tends to $\lambda\lt\infty$, then \[ q_k\to\frac{\lambda^k}{(1+\lambda)^{k+1}} \] (the right side is the geometric distribution).

Step 1: Part $1^{\circ}$ — counting

Fix a cell, say cell $1$, and require $m_1=k$. The other $r-k$ particles go into the other $n-1$ cells in any way. By stars and bars (Problem 58) with $r-k$ particles and $n-1$ cells, the number of ways is

\[ \binom{(n-1)+(r-k)-1}{r-k} = \binom{n+r-k-2}{r-k}. \]

Dividing by the number of all distributions:

\[ \boxed{q_k = \frac{\binom{n+r-k-2}{r-k}}{\binom{n+r-1}{r}}, \qquad k=0,1,\dots,r.} \]

Check: $\sum_k q_k = 1$, because $\sum_{k=0}^{r}\binom{n+r-k-2}{r-k} = \binom{n+r-1}{r}$ (the "hockey-stick" identity).

Step 2: Part $2^{\circ}$ — the ratio of consecutive terms

\[ \frac{q_{k+1}}{q_k} = \frac{\binom{n+r-k-3}{r-k-1}}{\binom{n+r-k-2}{r-k}} = \frac{r-k}{n+r-k-2}, \]

using $\binom{N}{j} = \frac{N}{j}\binom{N-1}{j-1}$ with $N=n+r-k-2$, $j=r-k$.

For $n\gt2$ the denominator is larger than the numerator ($n-2\gt0$), so $q_{k+1}\lt q_k$:

\[ \boxed{q_0 \gt q_1 \gt q_2 \gt \dots \gt q_r \qquad (n\ge3).} \]

Remark: for $n=2$ the ratio equals $1$ and all $q_k=\frac{1}{r+1}$ are equal (with two cells, every split $k, r-k$ is one distribution). So the strict inequalities need $n\ge3$ — which is always the case in physics, where $n$ is large.

Meaning: in Bose–Einstein statistics an empty cell is the most likely state of a given cell, and every additional particle makes the state less likely.

Step 3: Part $3^{\circ}$ — a product form

Writing the binomial coefficients with factorials and cancelling,

\[ q_k = \frac{n-1}{n+r-1}\cdot\frac{r}{n+r-2}\cdot\frac{r-1}{n+r-3}\cdots\frac{r-k+1}{n+r-k-1} = \frac{n-1}{n+r-1}\prod_{j=0}^{k-1}\frac{r-j}{n+r-2-j}. \]

(There are $k+1$ fractions; each numerator and denominator is a linear expression in $n$ and $r$.)

Step 4: Part $3^{\circ}$ — the limit

Divide numerator and denominator of each fraction by $n$ and use $r/n\to\lambda$, $1/n\to0$ (with $k$ fixed):

\[ \frac{n-1}{n+r-1} = \frac{1-\frac1n}{1+\frac rn-\frac1n}\ \to\ \frac{1}{1+\lambda}, \qquad \frac{r-j}{n+r-2-j} = \frac{\frac rn-\frac jn}{1+\frac rn-\frac{2+j}{n}}\ \to\ \frac{\lambda}{1+\lambda}. \]

A product of finitely many convergent factors converges to the product of the limits:

\[ \boxed{q_k \to \frac{1}{1+\lambda}\left(\frac{\lambda}{1+\lambda}\right)^k = \frac{\lambda^k}{(1+\lambda)^{k+1}}.} \]

This is a geometric distribution with ratio $\frac{\lambda}{1+\lambda}$; its terms add up to $1$ and its mean is $\lambda$, the average number of particles per cell.

Step 5: Numerical check ($\lambda = 2$)

As $n$ and $r$ grow with $r/n=2$, the exact values approach the geometric limit.