Counting, counting subsets , binomial theorem 
Can someone please help me? I have started by changing the form of each one for example k choose 2 as k!/(2!(k-2)!) and tried to show that all of them add up to n!/2!(n-2)! which is the same as n choose 2 but i got stuck. Is this the right approach? If so how do i simplify?

You are required to use a combinatorial argument to prove the identity.
Consider $n$ elements set $\{1, 2, 3, \cdots, n\}$ and we are to choose $2$ elements from this set.
One way to do this is $\binom{n}{2}$.
Another way, would be to count separately.
First, count the number of $2$-subsets where both elements are from $\{1, 2, \cdots, k\}$. This is $\binom{k}{2}$
Then, count the number of $2$-subsets where both elements are from $\{k + 1, k + 2, \cdots, n\}$. This is $\binom{n - k}{2}$
We not counting some $2$-subsets though. This is the one where one element is from $\{1, 2, \cdots, k\}$ and the other is from $\{k + 1, k + 2, \cdots, n\}$. Using multiplication principle, this is $k \cdot (n - k)$.
Hence, $$\binom{k}{2} + \binom{n - k}{2} + k(n - k) = \binom{n}{2}$$
If you are looking for an algebraic proof:
$$LHS = \binom{k}{2} + \binom{n - k}{2} + k(n - k)$$
$$=\frac{k(k - 1)}{2} + \frac{(n - k)(n - k - 1)}{2} + k(n - k)$$
$$=\frac{k^2 - k + n^2 - nk - n - kn + k^2 + k + 2kn - 2k^2}{2}$$
$$=\frac{n^2 - n}{2}$$
because the rest of the terms get cancelled out.
$$=\frac{n(n - 1)}{2}$$
$$=\binom{n}{2}$$
$$=RHS$$
