Choosing balls at random $a$ balls are chosen at random from a set of $n$ balls, and put in a bag. I would like to find the probability that $b$ given balls are in the bag. More specifically, the question is asking me to find this first directly, and then using the inclusion-exclusion principle.
Directly, I would say: there are $\binom{n}{a}$ ways of choosing the balls, and there are $\binom{a}{b}\binom{n-b}{a-b}$ ways of choosing $a$ balls which include the $b$. So the probability is:
$$\frac{\binom{a}{b}\binom{n-b}{a-b}}{\binom{n}{a}}$$
Is this right? To find an expression in terms of the inclusion exclusion principle, I have no clue - it would be great to get a hint. 
 A: Hints:  Assuming you chose balls $1, \cdots, b$, you have $\binom{n-b}{a-b}$ ways to select the remaining balls.
To use Inclusion-Exclusion, you can let $A_i$ be the ways of selecting n balls which do not include ball i, for $1\le i\le b$, and then find $P(\overline{A_1}\cap\cdots\cap\overline{A_b})$.
A: Let's call the set of all balls $S$, so that $S$ has cardinality $n$. Further, let's denote by $B$ the subset of interest whose cardinality is $b$.
I'm not certain about the calculation that there are 
$$
\begin{pmatrix}a \\b \end{pmatrix} \begin{pmatrix}n-a\\n-b \end{pmatrix}
$$
ways to choose $a$ balls that include $B$. I think of it this way: you're interested in subsets of size $a$ that contain $B$. These subsets are in bijection with subsets of $S \setminus B$ of size $a-b$. This is because given sets $B \subseteq A \subseteq N$, we can think of $A$ as the disjoint union $A = B \sqcup (A \setminus B)$. There may be a combinatorial identity that makes my numerator agree with yours, but I'm not sure there is.
Inclusion-exclusion is a real puzzler here. I'll continue thinking about it, but I'm not sure where to begin either.
