States with $\boldsymbol{n}$ Bose particles
Let’s extend our result to a situation in which there are $n$ particles present. We imagine the circumstance shown in Fig. 4–4. We have $n$ particles $a$, $b$, $c$, …, which are scattered and end up in the directions $1$, $2$, $3$, …, $n$. All $n$ directions are headed toward a small counter a long distance away. As in the last section, we choose to normalize all the amplitudes so that the probability that each particle acting alone would go into an element of surface $dS$ of the counter is \begin{equation*} \abs{\langle\quad\rangle}^2\,dS. \end{equation*}  First, let’s assume that the particles are all distinguishable; then the probability that $n$ particles will be counted together in $n$ different surface elements is \begin{equation} \label{Eq:III:4:15} \abs{a_1b_2c_3\dotsm}^2\,dS_1\,dS_2\,dS_3\dotsm \end{equation}  Again we take that the amplitudes don’t depend on where $dS$ is located in the counter (assumed small) and call them simply $a$, $b$, $c$, … The probability (4.15) becomes \begin{equation} \label{Eq:III:4:16} \abs{a}^2\abs{b}^2\abs{c}^2\dotsm dS_1\,dS_2\,dS_3\dotsm \end{equation}  Integrating each $dS$ over the surface $\Delta S$ of the counter, we have that $P_n(\text{different})$, the probability of counting $n$ different particles at once, is \begin{equation} \label{Eq:III:4:17} P_n(\text{different})=\abs{a}^2\abs{b}^2\abs{c}^2\dotsm(\Delta S)^n. \end{equation}  This is just the product of the probabilities for each particle to enter the counter separately. They all act independently—the probability for one to enter does not depend on how many others are also entering. Now suppose that all the particles are identical Bose particles. For each set of directions $1$, $2$, $3$, … there are many indistinguishable possibilities. If there were, for instance, just three particles, we would have the following possibilities: \begin{alignat*}{3} a\!&\to\!1&\quad a\!&\to\!1&\quad a\!&\to\!2\\ b\!&\to\!2&\quad b\!&\to\!3&\quad b\!&\to\!1\\ c\!&\to\!3&\quad c\!&\to\!2&\quad c\!&\to\!3\\[1ex] a\!&\to\!2&\quad a\!&\to\!3&\quad a\!&\to\!3\\ b\!&\to\!3&\quad b\!&\to\!1&\quad b\!&\to\!2\\ c\!&\to\!1&\quad c\!&\to\!2&\quad c\!&\to\!1 \end{alignat*}  There are six different combinations. With $n$ particles, there are $n!$ different, but indistinguishable, possibilities for which we must add amplitudes. The probability that $n$ particles will be counted in $n$ surface elements is then   \begin{align} \lvert a_1b_2c_3\dotsm&+a_1b_3c_2\dotsm+a_2b_1c_3\dotsm\notag\\ \label{Eq:III:4:18} &+a_2b_3c_1\dotsm+\text{etc.}+\text{etc.}\rvert^2 dS_1\,dS_2\,dS_3\dotsm dS_n. \end{align} 
 \begin{align} \lvert &a_1b_2c_3\dotsm+\,a_1b_3c_2\dotsm+a_2b_1c_3\dotsm\notag\\ \label{Eq:III:4:18} &+a_2b_3c_1\dotsm\!+\text{etc.}\!+\text{etc.}\rvert^2\, dS_1dS_2dS_3\!\dotsm dS_n. \end{align}   Once more we assume that all the directions are so close that we can set $a_1=$ $a_2=$ $\dotsb=$ $a_n=$ $a$, and similarly for $b$, $c$, …; the probability of (4.18) becomes \begin{equation} \label{Eq:III:4:19} \abs{n!abc\dotsm}^2dS_1\,dS_2\dotsm dS_n. \end{equation}   When we integrate each $dS$ over the area $\Delta S$ of the counter, each possible product of surface elements is counted $n!$ times; we correct for this by dividing by $n!$ and get \begin{equation} P_n(\text{Bose}) =\frac{1}{n!}\,\abs{n!abc\dotsm}^2(\Delta S)^n\notag \end{equation}  or \begin{equation} \label{Eq:III:4:20} P_n(\text{Bose}) =n!\,\abs{abc\dotsm}^2(\Delta S)^n. \end{equation}  Comparing this result with Eq. (4.17), we see that the probability of counting $n$ Bose particles together is $n!$ greater than we would calculate assuming that the particles were all distinguishable. We can summarize our result this way: \begin{equation} \label{Eq:III:4:21} P_n(\text{Bose})=n!\,P_n(\text{different}). \end{equation}  Thus, the probability in the Bose case is larger by $n!$ than you would calculate assuming that the particles acted independently. We can see better what this means if we ask the following question: What is the probability that a Bose particle will go into a particular state when there are already $n$ others present? Let’s call the newly added particle $w$. If we have $(n+1)$ particles, including $w$, Eq. (4.20) becomes \begin{equation} \label{Eq:III:4:22} P_{n+1}(\text{Bose})=(n+1)!\,\abs{abc\dotsm w}^2(\Delta S)^{n+1}. \end{equation}  We can write this as \begin{equation} P_{n+1}(\text{Bose})=\{(n+1)\abs{w}^2\,\Delta S\}n!\,\abs{abc\dotsm}^2(\Delta S)^n\notag \end{equation}  or \begin{equation} \label{Eq:III:4:23} P_{n+1}(\text{Bose})=(n+1)\abs{w}^2\,\Delta S\,P_n(\text{Bose}). \end{equation}   We can look at this result in the following way: The number $\abs{w}^2\,\Delta S$ is the probability for getting particle $w$ into the detector if no other particles were present; $P_n(\text{Bose})$ is the chance that there are already $n$ other Bose particles present. So Eq. (4.23) says that when there are $n$ other identical Bose particles present, the probability that one more particle will enter the same state is enhanced by the factor $(n+1)$. The probability of getting a boson, where there are already $n$, is $(n+1)$ times stronger than it would be if there were none before. The presence of the other particles increases the probability of getting one more.