The numbers of functions : There are not exist $f(i) < f(i+1) I solved this problem some days ago.  

Find the numbers of functions $f$ that satisfy these three conditions:
(1) $f$ is a bijection
  (2) $f : \{1, 2, 3,4\} \to \{1, 2, 3,4\}$
  (3)We do not have $f(i)<f(i+1)<f(i+2)$.

I found the number of such functions was $14$ , by counting one-by-one.
For other small numbers $n$ , I  know that it is the 'catalan numbers'.
I am very curious about these results, and am wondering:
How do I use some combinatorial method for general $n$?
Help me.
 A: You've stumbled upon a very interesting result! First of all, a bijection from $\{1,2,\dots,n\}$ to itself is what we call a permutation, I'll call such a function $\pi$. We can represent $\pi$ in one line form, meaning that we write 
$$
\pi=a_1,\dots,a_n
$$
where $a_i=\pi(i)$. We now can rewrite condition 3 by saying $\pi$ can have no subword $a_ia_ja_k$ that is strictly increasing. Here I allow subwords to be non-adjacent, as long as the order is maintained (i.e $i<j<k$). For instance, the permutation 
$$
4321
$$
satisfies our conditions, but 
$$
1324
$$
does not, as it contains the subword $124$ or also $134$. 
What we are working with is something called pattern avoidance in permutations. More can be seen here http://en.wikipedia.org/wiki/Permutation_pattern
In particular, you are counting the permutations which avoid the pattern $123$, in the sense I (somewhat loosely) described above. It is well known that these are counted by the Catalan numbers, as you suspected. For a rigorous bijective proof of this, see Permutations with Restricted Patterns and Dyck Paths by C. Krattenthaler, or many other sources. 
