How Many States Can N Particles in a Divided Box Assume?

[Fourier mn]

Homework Statement

N Particles in a box
A box consists of a gas that we'll treat as distinguishable particles. The box is divided into equal parts (right-left), by a partition with a small hole in it. Assume the particles are non-interacting and move back and forth between the two sides through a hole in a statistically independent fashion.
a. if each distinguishable particle is considered to have two states, L and R, depending on its position in side L or side R, what is the total number of different states of all the particles considered together?
b. for a given number of total particles N (with half in the right side and half in the left side), what is the multiplicity of states possible in terms of N, N right, N left?

so for part a. I think that the total number of states is (2)^N because each event is an independent event (each particle)-- like tossing a coin?! There are too states that all of the particles are in one of the sides (one left +one right).
For the multiplicity I got g= N!/(2(N/2)!). Are my assumptions correct?
I don't know how to do the second part.

I'll appreciate any suggestions,

Thank you,

Homework Equations

The Attempt at a Solution

 

[Jhero]

Hello,

For part a, your assumption is correct. Since each particle can be in either the left or right side, and there are N particles, the total number of states is 2^N.

For part b, the multiplicity can be calculated using the binomial coefficient formula, which is nCr = n!/(r!(n-r)!), where n is the total number of particles and r is the number of particles in one side. In this case, n=N and r=N/2. Therefore, the multiplicity is N!/(N/2)!(N/2)!, which simplifies to 2^N/N!.

Hope this helps! Let me know if you have any further questions.