Stirling number of the second kind and combinations The Stirling number of the second kind, denoted by $S(n,r)$, is defined as the number of $r$-partitions on a set of $n$ elements. Let $\binom{n}{r}$ (which is read $n$ choose $r$) denotes the number of $r$-element subsets of an $n$-element set. How to show that 
$$
\binom{n}{r}\leq S(n,r)
$$
for $2\leq r\leq n$.

We know that,
${n \choose r} = {n \choose {n-r}}$,
So it's sufficient to prove that,$${n \choose {n-r}} \le S(n,r)$$.
proof
For any $(n-r)$-subset of $n$ element set we can create (at-least one) $r$-partitions in the following way,
1) Put all the $(n-r)$ elements in a single partition.
2) partition the remaining(if any) $r$ elements into $r-1$ sets in $S(r,r-1)$ ways ($r \ge 1$).
Also, there are partitions which does not contain any  set of size $n-r$ if $r \notin \{1,n\}$.
By this you can conclude.
Now, check what happens when $r=2, n=4$. Is there any more like that? 
