Paths and combinatorics

Now I'll give a new perspective of some basic formulas in combinatorics:

Firstly ,let's find the number of paths from A(0,0) to B(a,b) ,where a path means moving to the right of the Ox axis and upper on the Oy axis.

There are $ \binom{a+b}{a}$ ways to reach point B(*)

Proof: let's consider a sequence of (a+b)numbers from the set {0,1} where we have a numbers of 0 and b numbers of 1 .

So we have to find the number of pairs$ (x_1,x_2,...,x_{a+b})$where $ x_i \in {{0,1}}$ which is $ \frac{(a+b)!}{a! \cdot b!}$

A little exercise would be to prove that $ \binom{n}{m}=\binom{n-1}{m}+\binom{n-1}{m-1}$

Well,consider O(0,0),A(n-m,m),B(n-m-1,m),C(n-m,m-1).

The number of paths to A is the number of paths which pass through B+no. of paths which pass through C and according to (*) we have the conclusion.

Other examples:1)$ \binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$

2)$ \sum_{a\leq c,b\leq d}\binom{a+b}{a} \cdot \binom{c+d-(a+b)}{c-a}=\binom{c+d}{d}$