Product of Combinations is Probability? For atomic orbitals:
E2 orb: $\binom{N-n_1}{n_2}$
E2 orb: $\binom{N-n_1-n_2}{n_3}$
E2 orb: $\binom{N-n_1-n_2-n_3}{n_4}$
...
En orb: $\binom{n_i}{n_i}$
now probability function is:
$P= N! \prod^{n}_{n=1}\frac{1}{n_{i}!}$
Why? In general?
[Update]
Every combination is greater than 1. So their product is greater than 1. How on earth can such multiplication lead to a probability function? Is the probability function scaled back to range $[0,1]$?
 A: As André Nicolas shows,
$$
\binom{N}{n_1}\binom{N-n_1}{n_2}\dots\binom{N-n_1-n_2-\dots-n_{k-1}}{n_k}=N!\prod_{i=1}^k\frac{1}{n_i!}
$$
This is an integer; it is the number of ways to arrange $N$ distinct things into $k$ bins with $n_i$ things in bin $i$.  If you sum this over all possibilities for the $\{n_i\}$, you get $k^N$ (the number of maps from $N$ things to $k$ bins).  So you would get a probability distribution if you divide by $k^N$.
The multinomial theorem states
$$
\left(\sum_{i=1}^kx_i\right)^N=\sum_{\sum_{i=1}^kn_i=N}N!\prod_{i=1}^k\frac{x_i^{n_i}}{n_i!}
$$
Setting $x_i=1$ for $1\le i\le k$, we get that
$$
k^N=\sum_{\sum_{i=1}^kn_i=N}N!\prod_{i=1}^k\frac{1}{n_i!}
$$
A: As you correctly point out, the terminology on page 10 of the lecture notes you linked to is incorrect: $P$ is the number of microstates making up the given macrostate; it is not a probability, since it typically exceeds one.
However, dividing $P$ by the total number of microstates $n^N$ does give the probability of the given macrostate, under the assumption that all microstates are a priori equally likely.  Since the constant $1/n^N$ is the same for all macrostates of the system, we may safely ignore it, as the author of the lecture notes does, when comparing the probabilities of different macrostates.
(If I were reviewing those notes, I'd also point out that the author seems rather excessively fond of the letter "n".  Having $n$ and $N$ as system parameters is confusing enough, but when he then introduces the macrostate parameters $n_1$, $n_2$ and up to $n_n$...)
A: Before the question is closed, I would like to give you a start.
Suppose $a+b+c+d=N$. Let us look at
$$\binom{N}{a}\binom{N-a}{b}\binom{N-a-b}{c}\binom{N-a-b-c}{d}$$
(the last term is $1$, it is just there to make things look nice.)
Calculate, using the usual formula for $\binom{n}{k}$.
The first term is
$$\frac{N!}{a!(N-a)!}.$$
The second term is 
$$\frac{(N-a)!}{b!(N-a-b)!}.$$
The third term is
$$\frac{(N-a-b)!}{c!(N-a-b-c)!}.$$
Note that $N-a-b-c=d$.
Multiply, and observe the very nice cancellations! We get
$$\frac{N!}{a!b!c!d!}.$$
The "general" case solution is basically the same, except that all those subscripts tend to make things less obvious.  
Added: By "the usual formula" for $\binom{n}{k}$ I mean
$$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$
Since the question has not yet been closed, I am adding a link to a Wikipedia entry which I think is quite well written, and which I hope you will give you all of the additional information you may need.
A: The "probability function" in the notes is usually referred to as the number of microstates $\Omega$ (I was taught this way). By the postulate of equal a priori probability, the probability, of a particular macrostate $i$ is $$p_i = \frac{1}{\Omega}$$So the probability function is different from the actual probability, which is probably what caused the confusion. 
Hence the $P$ in your notation is the number of ways $N$ particles can be classified into energy levels according to their energies (by the product rule, as your instructor and Andre above has done). Note that the analysis is classical as it assumes the particles filling the energy levels are distinct (come with labels).
You can find the same method in the Wikipedia article on Maxwell Boltzmann Distribution, except that it does not use confusing terminology of "Probability function". 
$$\Omega = N!\prod \frac{g_i^{N_i}}{N_i!}$$ Where on your case all $g_i = 1$ as there are no degenerate levels involved. Your instructor derives the MB distribution by trying to maximize the above number (as a system naturally seeks the maximum number of microstates) given the constraints (no exchange of particles or energy of the system with the surroundings) using Lagrange Multipliers. 
