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

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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.

Hint: Allow up to 3 particles per box and require at least 1 particle per occupied box to organize yourself. Here are some subscenarios:

As an easier example to work out for practice, show that for 3 indistinguishable particles in 4 distinguishable boxes, you would find

##### 1 Answer

I get

**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.

**All particles in the same box**

Here is the representative macrostate:

With

#9# boxes, you get#bb9# ways to have three-particle boxes.

**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.

**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.

**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.

Play around with this. You should get:

#a)# All particles in the same row:#3# configurationsHere 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.

Play around with this. You should get:

#a)# All particles in the same column:#3# configurationsHere 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

**METHOD TWO: DERIVING AN EQUATION**

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

They would each be separated by

#"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 **fixed** and the particles are **indistinguishable**.

To avoid double counting, we then divide by

This gives

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

when the *Statistical Mechanics* by Norman Davidson (pg. 66).

To check this equation, we put

#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

#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