Sets

We've presented here:http://theodormunteanu.blogspot.com/2011/07/paths-and-combinatorics.html
some sort of proof for a few identities:
There's a much larger spectrum of identities in combinatorics which can be deduced from the initial meaning of combinations:
for example:we must calculate the number of subsets with k elements from $\{a_1,a_2,...,a_n\}$which is the number of sets with k elements from $\{a_1,a_2,...,a_{n-1}\}$+the number of sets of (k-1)numbers from$\{a_1,...,a_{n-1}\}$and the kth element being $a_n$.From the definition of combination we have $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$
If we look further we will see that the same number of subsets is the no. of subsets with (k-2)elements from$\{a_1,a_2,...,a_{n-2}\}$ (the rest of 2 can be chosen from $\{a_{n-1},a_n}$)+$2\cdot$no.of sets with k-1 elements from $\{a_1,a_2,...,a_{n-2}\}$+no.of sets with k elements from$\{a_1,a_2,...,a_{n-2}}$.following the same reasoning we can conclude that
$\binom{m+n}{k}=\binom{m}{0}\cdot\binom{n}{k}+\binom{m}{1}\cdot\binom{n}{k-1}+...+\binom{m}{k}\cdot\binom{n}{0}${Try it)
Another easy application is to prove that $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n$ which that the number of subsets of $\{a_1,...,a_n}$ is the number of sets with 0,1,2,...elements.
try also to prove that:$\binom{n}{k}=\frac{n}{k}\cdot\binom{n-1}{k-1}$