A problem concerning catalan number Show that the number of ways to stack coins with n coins in the bottom row is denoted by Catalan number $$D_n = \frac{(_{2n}C_n)}{n+1}$$. I have tried in the way described in the picture below
 A: Add a row of $n+1$ coins below the bottom row of $n$ coins. Call the centre of the leftmost new coin $P$ and the centre of the rightmost new coin $Q$. The centres of all of the coins form a triangular lattice. We look at certain lattice paths from $P$ to $Q$. Specifically, we allow only moves one step up and to the right or one step down and to the right. Moreover, if it is possible to move up and to the right, we require that this move be made. It’s not hard to see that each stack of coins with an original bottom row of $n$ coins corresponds under these rules to a unique lattice path from $P$ to $Q$.
Each of these paths is essentially a Dyck path from $P$ to $Q$. Since each step takes it one unit to the right, it must have exactly $2n$ steps. It’s not hard to check that each Dyck path from $P$ to $Q$ corresponds to a stack of coins with an $n$-coin original base. Finally, it’s well-known that there are $C_n=\frac1{n+1}\binom{2n}n$ Dyck paths of length $2n$. Thus, $D_n=C_n$.
Example: The Dyck path shown below, with $n=4$, corresponds to the stack shown to its right:
         *   *                    O O  
        / \ / \                  O O O O
       *   *   *   *  
      /         \ / \  
     *   *   *   *   *  
     P               Q

