A beam of k particles is caught by n counters placed in a…

A beam of $k$ particles is caught by $n$ counters placed in a row, which register particles. Each particle can fall on any counter with equal probability. What is the probability that exactly $r$ counters register particles?

Step 1: All outcomes

Particle $1$ can hit any of $n$ counters, particle $2$ any of $n$, and so on. The particles are distinguishable (they arrive one by one), so

\[ |\Omega| = n^k. \]

Step 2: Choose which counters register

The $r$ registering counters can be chosen in $\binom nr$ ways. By symmetry each choice gives the same number of outcomes, so it is enough to count for one fixed set of $r$ counters.

Step 3: Fill $r$ fixed counters, none left empty

Count the ways to send $k$ particles into $r$ fixed counters so that every counter gets at least one. Let $E_i$ = "counter $i$ gets nothing". Without restriction there are $r^k$ ways. If $i$ given counters must stay empty, the particles use only $r-i$ counters: $(r-i)^k$ ways. By inclusion–exclusion,

\[ T(k,r) = r^k - \binom r1(r-1)^k + \binom r2(r-2)^k - \dots = \sum_{i=0}^{r}(-1)^i\binom ri(r-i)^k. \]

$T(k,r)$ is the number of onto maps from $k$ particles to $r$ counters. It equals $r!\,S(k,r)$, where $S(k,r)$ is the Stirling number of the second kind (the number of ways to split $k$ objects into $r$ non-empty groups).

Step 4: The probability

\[ \boxed{P_r = \frac{\binom nr}{n^k}\sum_{i=0}^{r}(-1)^i\binom ri(r-i)^k = \frac{\binom nr\,r!\,S(k,r)}{n^k}, \qquad 1\le r\le\min(k,n).} \]

Step 5: Example — $k=5$ particles, $n=4$ counters

$|\Omega|=4^5=1024$.

Most often $3$ of the $4$ counters register particles.

Step 6: Special cases

• $r=1$: all particles hit the same counter: $P_1 = n/n^k = n^{1-k}$.
• $r=k\le n$: every particle hits a different counter: $P_k = \binom nk k!/n^k = n(n-1)\cdots(n-k+1)/n^k$ (the birthday-problem formula, Problem 60).