# Thread: Show equality by combinational argument

1. ## Show equality by combinational argument

Hello,
I came across a problem, and I was confused on how to go about doing it, so was seeing if I could get some advice. I guess I'm a little confused to whats going on with it, and not sure how to show by combinational arguments, and algebraic manipulation. Any help is greatly appreciated, thanks!

2. Hello Mahonroy
Originally Posted by Mahonroy
Hello,
I came across a problem, and I was confused on how to go about doing it, so was seeing if I could get some advice. I guess I'm a little confused to whats going on with it, and not sure how to show by combinational arguments, and algebraic manipulation. Any help is greatly appreciated, thanks!
(a) $\displaystyle \binom{2n}{2}$ is the number of ways of choosing $\displaystyle 2$ items from a group containing $\displaystyle 2n$ items.

Now we may imagine the $\displaystyle 2n$ items to be divided arbitrarily into two groups of $\displaystyle n$ items each, and we may choose:
• $\displaystyle 2$ items from the first group (which can be done in $\displaystyle \binom{n}{2}$ ways); or

• $\displaystyle 2$ items from the second group (again in $\displaystyle \binom{n}{2}$ ways); or

• $\displaystyle 1$ item from each group (which can be done in $\displaystyle n \times n = n^2$ ways).

Thus the total number of ways of carrying out one of these three options is $\displaystyle \binom{n}{2}+\binom{n}{2}+n^2=2\binom{n}{2}+n^2$

Thus $\displaystyle \binom{2n}{2}=2\binom{n}{2}+n^2$

(b) $\displaystyle \binom{2n}{2} = \frac{2n!}{2!(2n-2)!}= \frac{2n(2n-1)}{2}= n(2n-1)$

$\displaystyle 2\binom{n}{2}+n^2=2\cdot\frac{n!}{2!(n-2)!}+n^2=\frac{2n(n-1)}{2}+n^2$

$\displaystyle = n(n-1)+n^2 = n(2n-1) = \binom{2n}{2}$