# How many different ways can you distribute three indistinguishable particles in a 3 x 3 ensemble of distinguishable boxes?

## Hint: Allow up to 3 particles per box and require at least 1 particle per occupied box to organize yourself. Here are some subscenarios: 1) All particles in the same box 2) Each particle in its own box 3) No two boxes in the same row contain the same number of particles. 4) No two boxes in the same column contain the same number of particles. 5) No two boxes contain the same number of particles. As an easier example to work out for practice, show that for 3 indistinguishable particles in 4 distinguishable boxes, you would find $20$ microstates.

Jul 19, 2017

I get $165$ using an equation I derived (which is correct), but when I do it the manual way I'm short by $36$... This question is too much to ask of you to do the manual way, particularly on an exam.

METHOD ONE: DRAWING OUT MACROSTATES

I'm going to do this a bit out of order, because having one particle per box is the most confusing one.

1) All particles in the same box

Here is the representative macrostate:

With $9$ boxes, you get $\boldsymbol{9}$ ways to have three-particle boxes.

3) Different boxes in the same row

Here is the representative macrostate:

With two particles in the same box, you have $6$ configurations of the remaining particle in its own box within the same row. With $3$ rows, that gives $\boldsymbol{18}$ configurations.

4) Different boxes in the same column

Here is the representative macrostate:

Rotate the box ${90}^{\circ}$, then repeat $\left(3\right)$ for $\boldsymbol{18}$ more, except we would have technically done it column-wise because the boxes are distinguishable.

5) Different boxes in different rows AND columns (at the same time)

Here is the representative macrostate:

You should get $4$ configurations with one two-particle box in the upper left, times three columns for the two-particle box within the same row equals $12$ configurations. Multiply by the three rows to get $\boldsymbol{36}$ configurations.

2i) Each particle in its own box, row-wise

Play around with this. You should get:

a) All particles in the same row: $3$ configurations

Here is the representative macrostate:

b) Two particles in the same row: $2$ configurations if two particles are in the first two columns, $2$ configurations if the particles are in columns $1$ and $3$, $2$ configurations if the particles are in the last two columns, times $3$ rows, for $18$ total.

Here is the representative macrostate:

c) All particles in different rows: $2$ diagonal configurations, $4$ configurations with two particles on the off-diagonal, for a total of $6$.

Here is the representative macrostate:

Apparently, I get $3 + 18 + 6 = \boldsymbol{27}$ here.

2ii) Each particle in its own box, column-wise

Play around with this. You should get:

a) All particles in the same column: $3$ configurations

Here is the representative macrostate:

b) Two particles in the same column: $2$ configurations if two particles are in the first two rows, $2$ configurations if the particles are in rows $1$ and $3$, $2$ configurations if the particles are in the last two rows, multiplied by the $3$ columns, for $18$ total.

Here is the representative macrostate:

c) All particles in different columns: We don't count these, because they are redundant.

Apparently, I get $3 + 18 = \boldsymbol{21}$ here.

However, that only accounts for $129$ microstates. That's not enough. Maybe I missed $36$ in $\left(5\right)$?

METHOD TWO: DERIVING AN EQUATION

We begin with the assumption of $N$ distinguishable particles in $g$ distinguishable boxes. If we arrange the boxes linearly, it doesn't change the number of microstates.

They would each be separated by $g - 1$ box walls, shown as $|$:

$\text{x x" cdots | "x x x" cdots | "x x} \cdots$

The walls are distinguishable, and thus could be swapped to obtain distinctly different configurations. With some amount of distinguishable things, there are a factorial number of ways to arrange them.

So, there are (g - 1 + N)! arrangements if the particles and boxes are both distinguishable. However, we wish to keep the box walls fixed and the particles are indistinguishable.

To avoid double counting, we then divide by N! and (g-1)! to account for redundant wall configurations and to ignore identical configurations of indistinguishable particles.

This gives

bb(t = ((g + N - 1)!)/(N!(g-1)!)" microstates")

when the $N$ particles are indistinguishable and the $g$ boxes are distinguishable. This matches the equation shown in Statistical Mechanics by Norman Davidson (pg. 66).

To check this equation, we put $\boldsymbol{3}$ indistinguishable particles in $\boldsymbol{4}$ distinguishable boxes to get:

color(blue)(t) = ((4 + 3 - 1)!)/(3! (4 - 1)!) = (6!)/(6 cdot 6) = (5! cdot cancel6)/(cancel6 cdot 6)

$= \frac{120}{6} = \textcolor{b l u e}{20}$ microstates color(blue)(sqrt""),

just like the example shows.

In your case, with $\boldsymbol{3}$ indistinguishable particles and $\boldsymbol{9}$ distinguishable boxes:

color(blue)(t) = ((9 + 3 - 1)!)/(3! (9 - 1)!) = (11!)/(6 cdot 8!) = (cancel(8!) cdot 9 cdot 10 cdot 11)/(6 cdot cancel(8!))

$= \textcolor{b l u e}{165}$ microstates

So we were short by $36$ in our manual guess...