TITLE: Fun with combinatorics and 80 business customers
QUESTION [6 upvotes]: In business with 80 workers, 7 of them are angry. If the business leader visits and picks 12 randomly, what is the probability of picking 12 where exactly 1 is angry? 
(7/80)(73/79)(72/78)(71/77)(70/76)(69/75)(68/74)(67/73)(66/72)(65/71)(64/70)*(63/69)*12=0.4134584151106464
What is the probability more than 2 are angry? 
My idea is to calculate the probability of 2,3,4,5,6, and 7 angry people just like did in the previous example and then add them altogether. 
In the previous example I can seat the one person 12 times. In all the different 12 spots, and then times by 12. The problem I have now is, how many times can I seat 2 people in 12 spots? If I use the combinatorics formula I will get a negative factorial.
There must be a much easier way than this.

REPLY [3 votes]: The denominators of the fractions stay constant. The total multiplication across the denominators is all the ways to pick 12 people from 80, where the order is retained:
$$ 80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74\cdot 73\cdot 72\cdot 71\cdot 70\cdot 69 = \frac{80!}{68!}
$$
We say that order is unimportant, so $\frac{80!}{68!\,12!} = {80 \choose 12}$ options
Then the numerators are the combination of the choices from the angry $(k)$ and non-angry $(12-k)$ groups, which are ${7 \choose k}$ and  ${73 \choose 12-k}$, so overall the probability is $$\frac{{7 \choose k}{73 \choose 12-k}}{80 \choose 12}$$
and checking this against your result for $k=1$ we have 
$$\frac{{7 \choose 1}{73 \choose 11}}{80 \choose 12} = \frac{68!\,12!}{80!}\frac{7!}{1!\,6!}\frac{73!}{ 62!\,11!} = \frac{12\cdot 7\cdot 68\cdot 67\cdot 66\cdot 65\cdot 64\cdot 63}{80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74} \approx 0.413458415
$$
The easiest of the possible calculations is where none of the chosen employees are angry $(k=0)$, which is 
$$\frac{{7 \choose 0}{73 \choose 12}}{80 \choose 12} = \frac{68!\,12!}{80!}\cdot 1 \cdot\frac{73!}{ 61!\,12!} = \frac{68\cdot 67\cdot 66\cdot 65\cdot 64\cdot 63\cdot 62}{80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74} \approx 0.305171687$$
Then all other cases can be worked similarly. 

You talk about getting a negative factorial when finding the combination of how to choose $2$ seats from $12$, but that is just ${12 \choose 2}= \frac {12!}{10!\,2!} = \frac{12\cdot 11}{2} = 66$ so I don't know how you arrived at a negative factorial.