Calculate probability using inclusion-exclusion and deduce formula for binomial coefficicient We choose uniformly a group of $k$ people selected from $n$. For $m \leq k$, calculate using inclusion-exclusion the probability that $m$ special people are in the group and then deduce that
\begin{align}
{n-m\choose k-m}=\sum_{j=0}^{m} (-1)^j {m\choose j} {n-j \choose k} 
\end{align}
Any suggestions? Thanks in advance!
 A: With this problem  the underlying poset consists  of nodes $Q\subseteq
[m]$ (the set of special people)  which represent groups of $k$ people
selected from the $n$ total where  the elements of $Q$ are missing, or
possibly more  of the special people.   The weights on the  $Q$ are as
usual $(-1)^{|Q|}.$ The  cardinality of the groups  represented at $Q$
is clearly ${n-|Q|\choose k}.$ Counting the groups presented at all $Q$
multiplied by their weight yields the closed form
$$\sum_{Q\subseteq [m]} (-1)^{|Q|} {n-|Q|\choose k} =
\sum_{q=0}^m {m\choose q} (-1)^q {n-q\choose k}.$$
On the other hand, counting  by computing the total weight contributed
by each of  the groups we find  that a group that has  all the special
people   only    appears   at   $Q=\emptyset$   with    total   weight
$(-1)^{|\emptyset|} = 1.$ A group that  has exactly $P$ of the special
people missing  where $|P|\ge 1$ appears  in all $Q\subseteq P$  for a
total weight of zero since
$$\sum_{Q\subseteq P} (-1)^{|Q|}
= \sum_{q=0}^{|P|} {|P|\choose q} (-1)^q = 0.$$
We conclude from these weights that the above sum counts exactly those
groups with  none of the  special people missing, these  having weight
one, and the others having weight zero. Therefore it is equal to
$${n-m\choose k-m}$$
Here we  have selected all $m$  special people first and  choose $k-m$
people from the remaining set.  Note.  When $n-|Q| \lt k$ the set of
groups represented at  $Q$ is empty, so that for  $m+k\gt n$ the nodes
of the top  $m+k-n$ rows of the poset represent  zero possible groups.
This  does not  affect  the calculation  of the  sum  of the  weights,
however, since a  group with the special people from  some $P$ missing
is represented by  the nodes of the embedded poset  spanned by the two
nodes $Q=P$ and $Q=\emptyset$ (which is entirely below $P$ rather than
above where  the potentially empty nodes  are).  None of the  nodes of
this poset are  empty if $n-|P|\ge k$ which is  a precondition for the
group to be realized in the first place.
 As a remark, observe that the sum is not difficult to evaluate. We
get
$$\sum_{q=0}^m {m\choose q} (-1)^q [z^k] (1+z)^{n-q}
= [z^k] (1+z)^n  \sum_{q=0}^m {m\choose q} (-1)^q  (1+z)^{-q}
\\ = [z^k] (1+z)^n \left(1-\frac{1}{1+z}\right)^m
= [z^k] z^m (1+z)^{n-m} = [z^{k-m}] (1+z)^{n-m}
\\ = {n-m\choose k-m}.$$
