Uniform sampling with replacement item frequency Suppose we are sampling from $N$ distinct items uniformly with replacement $M$ times. What can be said about the distribution of frequencies of items drawn? For example, if I sort all the frequencies from greatest to smallest and plot them, what will be the shape of the curve fitting this data?
For one particular item everything is more or less clear to me: expected number of times being drawn is $\frac{M}{N}$, variance is $\frac{M(N-1)}{N^2}$. But I don't understand how to get any general results.
EDIT: at the time I was writing the question I was just too confused about the whole experiment, so I couldn't formulate a precise question. After @Euxpraxis1981's answer and running some simulations myself here is what I am interested in.
Suppose $N \ll M$, we perform the described experiment and count how many items were drawn $0$ times, $1$ time, $2$ times, etc. If we plot these "item counts" for each "frequency", what will be the shape of the graph?
I tried with $N=100$ and different $M$ from $1000$ to $100000$. The histograms I got look a lot like binomial distribution pmf, with a mean (the greatest number of items being drawn) of $\frac{M}{N}$.
 A: Revised per OP revisions
Ok, I see what you are doing now :) Let's break it down:


*

*For a given item, the probability of choosing that item is $\frac{1}{N}$.

*If you perform M draws, the number of times you choose a particular item will have a $Bin(M,\frac{1}{N})$ distribution.

*However, you are constrained to have the total number of items drawn be M.


Item 3 normally precludes a simple analysis. However, you are doing something interesting, since you are effectively "anonymizing" the results (i.e., if an item were chosen 3 times in a particular run of the experiment, its irrelevant if it was item 1, 2, 3..etc)
Therefore, for each item $i$, we know that the number of times it could is chosen is $X_i\sim Bin(M,\frac{1}{N})$ (i.e., its marginal distribution is binomial, but the $X_i$ are NOT independent, so you cant just multiply the marginal probabilities.). 
Now, lets pick a particular integer-valued frequency, $0\leq f \leq M$. We can create indicator random variables $I_f(X_i)= 1$ $iff\;\; X_i=f$ and $0$ o/w.
Now, we have a precise statement of the question:
Let $F_f=\sum_{i=1}^N I_f(X_i)$ be the number of times an item was chosen $f$ times. We know that $\sum\limits_1^M F_i=N$ since we only have $N$ items and each item was either chosen $f$ times or not. Thus, the vector $F=(F_1,F_2...F_M)$ represents $N$ draws from a multinomial distribution defined by a vector of probabilities $p=(p_1,p_2,p_3...p_f)$.
What are these probabilities? Well, if $F_f=\sum I_f(X_i)$ then $E[F_f]=Np_f=\sum_{i=1}^N E[I_f(X_i)]$, thus, $p_f=\frac{\sum_{i=1}^N E[I_f(X_i)]}{N}=\frac{\sum_{i=1}^N P(X_i=f)}{N}=\frac{NP(X_i=f)}{N}=P(X_i=f) = Bin(f;M,\frac{1}{N})$
The step involving equality 3 is justified since each item has the same has the same probability of being selected on any given draw. Therefore, the resulting distribution will indeed be binomial, with expected value $\frac{M}{N}$
