• Support PF! Buy your school textbooks, materials and every day products Here!

Stat Mech : Balls in a box counting problem

  • Thread starter Opticmist
  • Start date
  • #1
3
0

Homework Statement



I already have the solution to the problem. Just need some help deciphering the logic behind it.

There are V balls, which are identical except for their color. N of them are blue and V-N are red. We place the balls inside a box so that V/2 balls are on each side.

How many arrangements(states) accommodate a p number of blue balls on the left hand side of the box.


Homework Equations



---

The Attempt at a Solution



Actual Solution:

[tex] \Gamma = \frac{N!}{p! (N-P)!} (\frac{V}{2})^p (\frac{V}{2})^{N-p} (V-N)! [/tex]

The explanation given for each term is as follows:

1) Pick p blue balls to occupy LHS
2) Spatially arrange p blue balls in LHS
3) Spatially arrange N-p blue balls on RHS
4) Spatially arrange all red balls inside the box.


Question:
I get the first three terms. I feel that the last term should not be there.

Since all red balls are identical, my feeling is that rearranging them in the remaining spots is not going to give any unique arrangements. Can someone please explain why the 4th term needs to be there?

Thanks!
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,187
The balls apparently aren't indistinguishable. If they were, then the first factor shouldn't be there either because any selection of p blue balls out of N would be the same.
 
  • #3
3
0
Makes perfect sense now. Thanks!
 

Related Threads for: Stat Mech : Balls in a box counting problem

  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
1
Views
2K
Top