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.

1 Answer
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.

DISCLAIMER: LONG ANSWER! (obviously)

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 bb9 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 bb18 configurations.

4) Different boxes in the same column

Here is the representative macrostate:

Rotate the box 90^@, then repeat (3) for bb18 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 bb36 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 = bb27 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 = bb21 here.

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

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 |:

"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 bb3 indistinguishable particles in bb4 distinguishable boxes to get:

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

= (120)/6 = color(blue)(20) microstates color(blue)(sqrt""),

just like the example shows.

In your case, with bb3 indistinguishable particles and bb9 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!))

= color(blue)(165) microstates

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