Given n particles, each of which can be found with equal…

Given $n$ particles, each of which can be found with equal probability $\frac1N$ in each of $N$ $(N\gt n)$ cells. Find the probability that:
$1^{\circ}$ exactly one particle falls into each of $n$ specified cells;
$2^{\circ}$ exactly one particle falls into each of some $n$ cells.

Step 1: All outcomes

Particle $1$ can go to any of $N$ cells, particle $2$ to any of $N$ cells, and so on:

\[ N\cdot N\cdots N = N^n \]

equally likely placements.

Step 2: $1^{\circ}$ — one particle in each of $n$ specified cells

The $n$ particles must occupy the $n$ given cells, one per cell. This is a way of assigning the particles to the cells one-to-one: $n!$ ways.

\[ \boxed{P_1 = \frac{n!}{N^n}.} \]

Step 3: $2^{\circ}$ — one particle in each of some $n$ cells

Now the occupied cells may be any $n$ of the $N$ cells: choose them in $\binom Nn$ ways, then place the particles in $n!$ ways:

\[ \boxed{P_2 = \binom Nn\frac{n!}{N^n} = \frac{N!}{(N-n)!\,N^n} = \frac{N(N-1)\cdots(N-n+1)}{N^n}.} \]

In other words: no two particles share a cell. Placing the particles one by one: the first may go anywhere, the second must avoid one cell ($\frac{N-1}{N}$), the third must avoid two cells ($\frac{N-2}{N}$), and so on:

\[ P_2 = 1\cdot\frac{N-1}{N}\cdot\frac{N-2}{N}\cdots\frac{N-n+1}{N} = \prod_{j=0}^{n-1}\left(1-\frac jN\right). \]

Step 4: Example — the birthday problem

Take $N=365$ "cells" (days of the year) and $n$ "particles" (people). $P_2$ is the probability that all birthdays are different:

With just $23$ people a shared birthday is more likely than not.

Step 5: Physical meaning

This is the counting of Maxwell–Boltzmann statistics (distinguishable particles). Compare with Bose–Einstein statistics in Problem 58, where the particles are indistinguishable and the probabilities are different.