Statistical Mechanics: Computing a system's microstate multiplicity I have a general question concerning how to compute the microstate multiplicity of a system, in my lecture notes, for a system of $N$ weakly coupled oscillator and $Q$ energy quantas, the multiplicity is given by:
$\Omega(N,Q)= \binom{N+Q-1}{Q}$
whereas for another system of N slots and Q samples, the system's multiplicity is simply computed by knowing how many ways can the Q samples be fitted in the N slots:
$\Omega=\binom{N}{Q}$
I don't really understand the difference between these two situations,
Thank your for your answer,
 A: In the second case
$$\Omega = \binom{N}{Q}$$ 
you're picking $Q$ distinguishable balls out of a bag of $N$ and putting each into one of $Q$ cups lined up in a row, but it doesn't matter how they're ordered in the cups. 
In the first case
$$\Omega = \binom{N+Q-1}{Q}$$
you're taking $Q$ indistinguishable balls and putting them into one of $N$ cups, and there can be any number of them in each cup.
It just so happens that you can think of the second problem in terms of binomial coefficients by re-imagining the scenario using a technique called stars and bars, but it should be clear from the descriptions I just gave that they are distinct counting problems.
Addendum. Intuition
Consider the case $N = 2$ and $Q = 2$.  
In the second case, this would correspond to having a bag containing a red ball and a blue ball, taking these balls from the bag, and putting each one in one of two cups lined up on a table but considering any ordering of the balls in the cups to be the same.  We can see in this case that only one result is obtained: the red ball goes in one of the cups, and the blue ball goes in the other one.  There is one configuration in total:
$$\Omega = \binom{2}{2} = \frac{2!}{2!1!} = 1$$
In the first case, this would correspond to having a bag containing two red balls, and two cups lined up on the table, but this time, we can put any number of red balls in either of the cups, and the resulting configuration is reported as the number of balls in the first cup along with the number of balls in the second cup.  There can be two balls in the first and none in the second, one in the first and one in the second, or none in the first and two in the second.  Three configurations in total.
$$\Omega = \binom{3}{2} = \frac{3!}{2!1!} = 3$$
The reader might consider trying other cases with small $N$ and $Q$ to develop her intuition further.
