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?
 A: 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.
$$=\frac{n(n - 1)}{2}$$
$$=\binom{n}{2}$$
$$=RHS$$
A: On a $2,k$ lattice, the left hand side counts the number of paths that pass through $(2,k)$, plus the number of paths that pass through $(0,k)$, plus the number of paths that pass through $(1,k)$ (as $\binom{k}{1}=k$ and $\binom{n-k}{1}=n-k$).
A: 
Combinatorically prove that 
$~\displaystyle \binom{k}{2} + \binom{n-k}{2} + [k(n-k)] = \binom{n}{2} ~: ~2 \leq k \leq n.$

RHS is number of different ways of selecting $2$ items, without replacement, from $n$ items.
Assume that before these items are selected, any fixed subset of $k$ distinct items is formed.
Then, when selecting these $2$ items, there are $3$ disjoint possibilities:

*

*Both of the $2$ items come from the (fixed) subset of $k$ items:
Number of ways that this can occur is $~\displaystyle \binom{k}{2}.$

*Both of the $2$ items come from the (fixed) complementary subset of $(n-k)$ items:
Number of ways that this can occur is $~\displaystyle \binom{n-k}{2}.$

*One of the $2$ items come from the (fixed) subset of $(k)$ items, and one of the $2$ items comes from the fixed complementary subset of $(n-k)$ items:
Number of ways that this can occur is $~\displaystyle \binom{k}{1} \times \binom{n-k}{1} = k(n-k).$
Note that the $(3)$ LHS terms exactly correspond to the above $(3)$ computations.  Further these computations (in effect) partition all of the possible subsets of $(2)$ distinct items, selected from the $(n)$ items.
This means that the computations represent disjoint collections of subsets of $(2)$ items, where the union of these $(3)$ collections equals the collection of all possible subsets of $(2)$ distinct items, selected from the $(n)$ items.
Therefore, from a combinatoric perspective, the sum of the $(3)$ LHS terms must equal the RHS.
