# Show equality by combinational argument

• Sep 30th 2009, 07:10 AM
Mahonroy
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!
• Sep 30th 2009, 10:34 AM
Hello Mahonroy
Quote:

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

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

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