Take seven courses out of 20 with requirement 
To fulfill the requirements for a certain degree, a student can choose to take any 7 out of a list of 20 courses, with the constraint that at least 1 of 7 courses must be a statistics course. Suppose that 5 of the 20 courses are statistics courses.
From Introduction to Probability, Blitzstein, Hwang

Why is ${5 \choose 1}{19 \choose 6}$ not the correct answer?
 A: Because you've double counted, you could choose the same two stat classes in two different ways.  If you choose stat class 1 out of five and then stat class 2 out of the remaining 19, that's going to happen again when you choose stat class 2 out of the five and stat class 1 out of the remaining 19.
A: There are $\binom{20}{7}$ possibilities, but $\binom{15}{7}$ of them are not ok, hence the answer is $\binom{20}{7}-\binom{15}{7}$.
A: The given answers are all correct. Nevertheless, there was one argument which made it very clear (to me) that 
$$
{5 \choose 1} {19 \choose 6}
$$
is not the correct answer. It goes as follows:
Question: Suppose there are 20 different courses and five of those focus on statistics. You have to choose one "major" and six side courses. How many different combinations are possible, if the "major" must focus on statistics?  
Answer: We have to choose one out of the five statistics courses and six out of the remaining 19 courses. Thus, the answer is 
$
{5 \choose 1} {19 \choose 6}
$. 
However, in the original question asked by Blitzstein, there is no reference to a "major" course. Thus, the original question (without the major course) has fewer arrangements. 
A: The animated video from the stat110x course gives a clear picture of why Vandermonde's Identity instead of $\binom{5}{1} · \binom{19}{6}$ is used. 
Vandermonde's Identity: 
$ \binom{m+n}{r} = \sum_{k=0}^r \binom{m}{k} \binom{n}{r-k} $
Using the video as reference for the stat class question, $\binom{m}{k}$ represents the stat classes and $\binom{n}{r-k}$ represents the none stat classes. For every stat class chosen, the number of $r$ choices reduces, hence, r-k from when choosing the none stat classes.
$ \binom{5+15}{7} - \binom{5}{0}\binom{15}{7} = \binom{20}{7} - \binom{15}{7}$
We minus $\binom{5}{0}\binom{15}{7}$ because this represents the permutations of choosing 0 stat class and 7 none stat classes.
