Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$ So i was given this question. Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$ 
Here is my solution:
Given $2n$ objects, split them into $2$ groups of $n$, $A$ and $B$. $2$-combinations can either be assembled both from $A$, both from $B$ or one from each. There are ${n \choose 2}$ from $A$ and ${n \choose 2}$ from $B$. For the mixed pair, each choice from $A$ can be coupled with $n$ choices from $B$ so the total is $n^2$. 
Therefore ${2n\choose 2} = 2{n \choose 2} + n^2$ 
Is this correct. Is there any other combinatoric proof to solve this?
 A: There is a generalization of this fact that I cannot resist sharing. Here is how I came across it: Consider the scenario where you keep additively splitting a positive integer into two positive integers as a tree diagram, until you are left with only $1$'s as the leaves. Then you take the product of the two numbers at each branching and sum all such products. Remarkably, applying this process to $n$ will always yield $\binom{n}{2}$ as the final number. Try it with $8$ and you get $28$ every time. The inductive step of the strong induction argument for this fact uses the fact that, if you split $n=a+b$ then $$\binom{a}{2}+\binom{b}{2}+ab=\binom{a+b}{2}=\binom{n}{2}.$$ A combinatorial proof double counts the number of ways of choosing two people out of $n=a+b$ people. We can either choose $2$ out of the $a$ or $2$ out of the $b$ or $1$ from each. The original post asks for a combinatorial proof of this fact when $a=b.$
A: Just another way to think about it:
Let's find the coefficient of $x^2$ in the expansion of $(1+x)^{2n}$. On one side, it's equal to ${2n}\choose2$. On the other hand, it's equal to the coefficient of $x^2$ in the expansion of $(1+x)^n(1+x)^n$ which is equal to the sum of products of coefficients of $x^0$ and $x^2$, $x^1$ and $x^1$, and $x^2$ and $x^0$ in the expansion of $(1+x)^n$. That's equal to ${n\choose2}\cdot{ n\choose0}+{n\choose1}\cdot{ n\choose1}+{n\choose0}\cdot{ n\choose2}$ and the result follows.
A: A more direct (and, in my opinion, simple) approach is to expand and simplify until both sides are equivalent.
$${{2n}\choose{2}}=2{{n}\choose{2}}+n^2$$
First, expand both sides using ${{n}\choose{k}}=\frac{n!}{k!(n-k)!}$:
$$\frac{(2n)!}{2!(2n-2)!}=2\left[{\frac{n!}{2!(n-2)!}}\right]+n^2$$
Expand again, this time focusing on the factorials:
$$\frac{(1)(2)(3)...(2n-2)(2n-1)(2n)}{(1)(2)*(1)(2)(3)...(2n-2)}=2\left[\frac{(1)(2)(3)...(n-2)(n-1)(n)}{(1)(2)*(1)(2)(3)...(n-2)}\right]+n^2$$
In this format, it's easy to see that some variables of the factorials on each side cancel each other out:
$$\frac{(2n-1)(2n)}{2}=(n-1)(n)+n^2$$
Factor in the monomials on each side:
$$\frac{4n^2-2n}{2}=n^2-n+n^2$$
Simplify both sides for the last time:
$$2n^2-n=2n^2-n$$
This method takes a little longer but I feel it's more intuitive and solid.
A: Theorem: C(2n, 2) = 2C(n, 2) + n^2
Proof.
Suppose we have 2 groups, both of size n, which combine to make a class of size 2n, and we wish to know the number of ways we can choose 2 members from the class.  This is represented by C(2n, 2).
There are C(n, 2) ways we can choose 2 students from a single group. Since there are 2 groups, there are 2C(n, 2) ways we can choose 2 students from a single group.  Finally, there are n^2 ways to choose to members from the class where each member is from a different group, so there are 2C(n, 2) + n^2 ways to choose 2 members from the class of size 2n.
Therefore C(2n, 2) = 2C(n, 2) + n^2∎
