How many ways can you be diagnosed with a Substance Use Disorder? In the Diagnostic and Statistical Manual for Mental Disorders, Fifth Edition (DSM-5), one can be diagnosed with a Substance Use Disorder, e.g., Cannabis Use Disorder, if $2$ (or more) of $12$ possible symptoms have been present for $12$ months or more, and if those symptoms cause "clinically significant impairment or distress".
Concentrating only on the "$2$ of $12$ symptoms" aspect, how many different symptom combinations exist such that one could be diagnosed with a Substance Use Disorder?
This question is similar to In how many ways can you paint 90 distinct buckets?, How many ways “ARRANGE” can be arranged? [closed], and some other questions, but not similar enough for me to figure out the answer to my question.
There is an academic article that provides answers to similar symptom combination questions:

Galatzer-Levy, Isaac R., and Richard A. Bryant. "636,120 ways to have posttraumatic stress disorder." Perspectives on Psychological Science 8, no. 6 (2013): 651-662.

The authors use "an $n$ choose $k$ binomial equation with replacement", but I do not understand how to calculate an answer using the formula. (I tried several times, but I could not duplicate their results.) Part of the formula is:
$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
For starters, I'm not sure what $k$ would be for the question I am posing. I thought it would be $k=2$, but I'm not sure. 
The complete formula is on page 655 of the article (PDF). It is:
$$\prod_{n = i}\left[\sum \binom{n}{k}\right],~\text{where}~\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
Thanks! ~ Mark
 A: 
In the Diagnostic and Statistical Manual for Mental Disorders, Fifth Edition (DSM-5), one can be diagnosed with a Substance Use Disorder, e.g., Cannabis Use Disorder, if $2$ (or more) of $12$ possible symptoms have been present for 12 months or more, and if those symptoms cause "clinically significant impairment or distress".
Concentrating only on the "$2$ of $12$ symptoms" aspect, how many different symptom combinations exist such that one could be diagnosed with a Substance Use Disorder?

As you may know, $n!$ is the product of the first $n$ positive integers if $n$ is a positive integer.  The number $0!$ is defined to be $1$.  Also, the formula
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
counts the number of subsets of size $k$ of a set with size $n$, where $\binom{n}{k}$ is read $n$ choose $k$.
To give an example, suppose we wish to find the number of subsets of size $2$ that can be drawn from the five-element set $S = \{a, b, c, d, e\}$.  There are five ways of selecting the first element, four ways of selecting the next element, and three ways of selecting the third element.  Thus, at first glance, there would appear to be $5 \cdot 4 \cdot 3 = 60$ such subsets.  However, selecting $c$, then $a$, then $b$ results in the same subset as selecting $a$, then $b$, then $c$.  How many ways could we draw the same three elements?  Say we select $a$, $b$, and $c$.  There are three ways to select one of these elements first, two ways to select one of the remaining two elements second, and one way to select the remaining element third.  Thus, our initial count was too large by a factor of $3! = 3 \cdot 2 \cdot 1 = 6$ since there were six different orders in which we could have chosen the same three elements.  Thus, there are actually only
$$\frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1} = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 2 \cdot 1} = \frac{5!}{3!2!} = 10$$
such subsets.  They are:
$$\{a, b, c\}, \{a, b, d\}, \{a, b, e\}, \{a, c, d\}, \{a, c, e\}, \{a, d, e\}, \{b, c, d\}, \{b, c, e\}, \{b, d, e\}, \{c, d, e\}$$
Since a diagnosis of Cannabis Use Disorder requires the presence of two or more of twelve possible symptoms in a patient, the number of possible clusters of symptoms that could result in that diagnosis for a single patient is
\begin{align*}
\sum_{k = 2}^{12} \binom{12}{k} & = \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} + \binom{12}{6}\\
 & \quad + \binom{12}{7} + \binom{12}{8} + \binom{12}{9} + \binom{12}{10} + \binom{12}{11} + \binom{12}{12}\\
& = 66 + 220 + 495 + 792 + 924 + 792 + 495 + 220 + 66 + 12 + 1\\
& = 4083
\end{align*}
Here is another way to arrive at the same result:  Each of the twelve symptoms could be present in a patient or not. Thus, there are two possibilities for each symptom, or $2^{12}$ possible subsets of symptoms.  A patient should not be diagnosed with Cannabis Use Disorder if none or only one of the symptoms are present.  Therefore, the number of possible ways cannabis use disorder could be diagnosed in a single patient is
$$2^{12} - \binom{12}{0} - \binom{12}{1} = 4096 - 12 - 1 = 4083$$
In my reading of the article (I do not have a background in psychology or psychiatry), the formula on page 655 of the article by Galatzer-Levy and Bryant applies when there are multiple symptom criteria that could be met for a diagnosis such as Post-Traumatic Stress Disorder (which, if I understand the article correctly, covers a spectrum of disorders).  In that case, we have to multiply our results for each set of symptom criteria in order to calculate the total number of ways PTSD could be diagnosed.
