A sample of 10 articles was chosen out of 20. What is the probability of a certain item being among them? 
A random sample of 10 items was chosen from 20. What is the
  probability for a certain article to be among the sample of 10?

I tried to do this:
$$\frac{?}{20\choose{10}} $$
I'm having trouble with the denominator. My professor solved it like this:
$$p(\lnot A) = \frac{19\choose{10}}{20\choose 10} = 1/2$$
$$p(a) = 1- p(\lnot A) = 1/2$$
I have two questions: 1) how can I solve it how I was doing initially and 2) what did my professor do exactly?
 A: In your denominator, you have counted the total number of ways to select $10$ distinct items out of a group of $20$ items, such that the order of selection is not relevant.
Therefore, in your numerator, you should count the total number of such selections in which a specific single item is included among the $10$ selected.  To do this, imagine that you have "pre-selected" that item.  Now, there are $19$ items remaining, and you have $9$ more items to select.  Thus, there are $$\binom{19}{9}$$ such desired outcomes, out of the possible $\binom{20}{10}$.  Note that it is important to keep the way you counted such selections consistent when enumerating the desired outcomes versus the possible outcomes--otherwise, you're not counting the same things.
The way the professor solved the question is similar but slightly different; this approach instead counts the complementary outcome of avoiding the selection of the desired object.  To do this, imagine simply throwing that item away.  Now you have $19$ items from which $10$ must be selected, giving $$\binom{19}{10}$$ choices.  Then the desired probability is $1$ minus the complementary probability.

If all of the above is difficult to conceptualize, suppose you have $20$ balls numbered $1$ through $20$ inclusive.  Suppose all of the balls are white, except for ball number $8$, which is black.  If you want to count the number of ways to select $10$ of the balls such that one of them is the black $8$-ball, then since it doesn't matter the order in which you select the balls, just pick the $8$-ball first, leaving you with $9$ more balls to choose from the $19$ remaining.  Conversely, with the professor's approach, you simply discard the $8$-ball, leaving you with $19$ balls from which you must select $10$.
Incidentally, the fact that the resulting probability is $1/2$ also proves that $$\binom{19}{9} = \binom{19}{10},$$ which we also see because $$\binom{n+m}{n} = \frac{(n+m)!}{n! m!} = \binom{n+m}{m}.$$
A: Let the specific item you want in the sample be called $A$. What are the possible combinations of $10$ items  containing $A$? Since one of the $10$ items is known to be $A$, you only need to consider possible combinations of the rest of $9$ which is $\binom{19}{9}$. 
$19$ because it does not include $A$; $A$ is already in there.  
Total combinations is indeed $\binom{20}{10}$. Number of combinations containing $A$ is $\binom{19}{9}$ so the probability is: $$\frac{ \binom{19}{9}}{\binom{20}{10}}$$
Your professor simply considered combinations NOT containing $A$ which are  $\binom{19}{10}$ because $A$ is not included. Hence, probability of not containing $A$ is $$\frac{\binom{19}{10}}{\binom{20}{10}}$$
so that $$1 - \frac{\binom{19}{10}}{\binom{20}{10}} = \frac{1}{2} = \frac{\binom{19}{9}}{\binom{20}{10}} $$
