Math

This is really pissing me off :hmmm:
I was given a take home quiz for my discrete math class and there is this problem that we didn’t even talk about on there. I’ve read the single example (meaning that that is all the info given in the book about this sort of this) of that type of problem and I can’t possibly think of how to use this method of proof to solve the problem given to me.

“Convince yourself that the following statement is true, but use a combinatorial argument to verify it validity.

(2n)C(2) = 2(n)C(2)+n^2

Or if you like it this way,

C(2n,2) = 2 C(n,2)+n^2

I proved it algebraically…I know it’s true

I’m thinking of just saying:
“Choosing 2 from a set of 2n elements is the same as; twice the number of choosing 2 from a set of n elements and then adding n squared more to that.
:ph34r:

revision:

Let n be positive and an integer.
Suppose S is a set with 2n elements.
Then numbers of subsets of S of size 2 can be calculated by thinking of S as consisting of two pieces:
One with n elements {x(1), x(2), …, x(n)}
The other with a second set of n elements {x(n+1), x(n+2), …, x(2n)}

Any subset of S with 2 elements either contains the second set of n elements {x(n+1), x(n+2), …, x(2n)} or not
If it does contain {x(n+1), x(n+2), …, x(2n)}, then it contains 2 elements from set {x(1), x(2), …, x(n)}, times 2
If it does not contain {x(n+1), x(n+2), …, x(2n)}, then it contains n squared elements.
I was thinking about it too hard. I wanted it to be more like a proof... and complicated and stuff...Owell...

revision version 2.0:

(Left Hand Side)
Lets just say that you have a bin with 2n number of things in it, and you pick 2 items from that bit. This is C(2n,2)

(Right Hand Side)
The second way of looking at this is that you have two bins with n number of things in them; you want to pick two items (total) from these bins. This is broken into three cases:
Case 1) Both items are picked from the first bin (of n things). This is C(n,2)
Case 2) Both items are picked from the second bin. This is C(n,2)
Case 3) one item is taken from each of the bins. One item from the first bin, C(n,1); and one item from the second bin, C(n,1). This is C(n,1) * C(n,1). C(n,1)=n. So in total for this case is n^2

Adding these cases up gives C(n,2) + C(n,2) + n^2
Which can be simplified to 2C(n,2)+ n^2

This is what I was to prove… that C(2n,2) = 2C(n,2)+ n^2