Statistical count I am reading the book"Heat and Thermodynamics" by Mark Waldo Zemansky and Richard Dittman. In the chapter "Statistical Mechanics" it says if I have $N_{i}$ distinguishable particles in any of $g_{i}$ degenerate states associated with energy $\epsilon_{i}$ and if there is no restriction on multiple occupancy then there will ${(g_i)}^{N_i}$ number of ways in which those particles can be occupied. Then it says in case of indistinguishable particles the number of ways become $\dfrac{{(g_i)}^{N_i}}{N_{i}!}$.  
when I do it in tabular method with $g_{i}$=3 and $N_{i}$=3 with the indistinguishable particles it gives 10 ways, not in agreement with what's written. Please help!
 A: I think the following was the argument that the authors used to arrive at their expression:

The first particle can choose to be in any one of the g states; the
  second again. So N particles have a total of g*g*g*... choices - which
  is the equation given. When they are indistinguishable you divide by N
  factorial since that is how many times each combination shows up.

However, I come up with a different answer. Read on.
Your tabular example (3 distinguishable particles, 3 states) should look like this:
state |  1  |  2  |  3  |
------+-----+-----+-----+
 1    | 123 |     |     |
 2    | 12  |   3 |     |
 3    | 12  |     |   3 |
 4    | 1 3 |  2  |     |
 5    | 1   |  23 |     |
 6    | 1   |  2  |   3 |
 7    | 1 3 |     |  2  |
 8    | 1   |   3 |  2  |
 9    | 1   |     |  23 |
10    |  23 | 1   |     |
11    |  2  | 1 3 |     |
12    |  2  | 1   | 1 3 |
13    |   3 | 1   |     |
14    |     | 1 3 |   3 |
15    |     | 1   | 1 3 |
16    |   3 | 1   |     |
17    |     | 1 3 |     |
18    |     | 1   |   3 |
19    |  23 |     | 1   |
20    |  2  |   3 | 1   |
21    |  2  |     | 1 3 |
22    |   3 |  2  | 1   |
23    |     |  23 | 1   |
24    |     |  2  | 1 3 | 
25    |   3 |     | 12  | 
26    |     |   3 | 12  | 
27    |     |     | 123 | 

27 states with distinguishable particles. Now for a small sample, as you pointed out, dividing the number of states by N! to compute the "indistinguishable" case doesn't make sense - it doesn't even lead to a whole number.
The "correct" answer for the second part is really the answer to the question: "in how many ways can we partition N identical particles with (g-1) partitions? Why (g-1)? Because one we put all the particles in a line, we need only (g-1) partitions to generate all possible states.
So there are (N+g) places to put (g-1) partitions. This we can solve - the answer is
$$\frac{(N+g)!}{(N+g-(g-1))!\cdot(g-1)!}$$
When N is very large (much larger than g), this becomes approximately
$$\frac{N^g}{g!}$$
Huh. That's funny. Now it's $$N^g$$ where before it was $$g^N$$.
Just to confirm the math - for our 3 particles, 3 states example, the equation becomes
$$\frac{5!}{3!\cdot2!}= \frac{120}{6\cdot2} = 10$$
as expected.
I have to conclude that the book is wrong... it wouldn't be the first time.
Clarification:
I realize that the "how do you partition N particles into g states" formula isn't very intuitive - so let me elaborate a little bit.  Here is one way to think about the problem. Draw N+g-1 boxes in a row. Now place g-1 check marks in arbitrary boxes (at most one per box). Starting from the left, count the number of "open" boxes. Each time you hit a marker, you reset your counter and start again. Since you had N+g-1 boxes with g-1 check marks, you will have reset your counter g-1 times, and counted a total of N empty boxes (and placed these into g counters). Thus, this method gives you a possible partition of N into g states. And there are (N+g-1)choose(g-1) different ways to do this.  
You can see the above equation mentioned (as equation (3)) in the paper http://www.linux.bucknell.edu/physics/ligare/hoStatMech.pdf
