Talk:Combinatorics/Subsets of a set-The Binomial Coefficient

Hi, I am trying to understand permutations and combinations, and since I am not absolute certain I would like to ask before any editing is taking place.

In the proof: "Proof: Let X=\{1,2\cdots n\}. These numbers may be listed in various orders, called permutations. There are n! of these permutations because the first term may be any of the n numbers, the second any of the remaining n-1 numbers and so on.

Let us count these permutations in a different way now. Then we shall apply double counting as described in the previous chapter.

Let Y be a subset of X having k elements. Then there are k! permutations of the elements of Y. Similarly there are (n-k)! permutations of the elements not in Y. If we attach any one of these (n-k)! permutations to the right end of any one of the k! previous permutations, the ordered sequence of n elements thus obtained is one of the permutations of X. To get all the permutations of X we repeat the procedure with Y replaced by each of the k-order subsets. Thus the total possible permutations would be T.n!(n-k)! where T is the number of k-order subsets. That is because total permutations = adding n!(n-k)! the number of times equal to the number of k-order subsets = T.n!(n-k)!. Now this number must be equal to n! and so we have T = The number of k-order subsets of X = \frac{n!}{k!(n-k)!} = {n\choose k}. Q.E.D."

It says there are n!(n-k)! total permutations of X

"If we attach any one of these (n-k)! permutations to the right end of any one of the k! previous permutations, the ordered sequence of n elements thus obtained is one of the permutations of X"

And

"Thus the total possible permutations would be T.n!(n-k)! where T is the number of k-order subsets"

Should it not be T.k!(n-k)! of k order subsets there exists?

I mean there are k! permutations of Y, and (n-k)! permutations of X without Y, so there should be T.k!(n-k!) of X, which also equals n!, so T. = n! / (k!(n-k)!)

However as it says now in the ending:

"That is because total permutations = adding n!(n-k)! the number of times equal to the number of k-order subsets = T.n!(n-k)!. Now this number must be equal to n! and so we have T = The number of k-order subsets of X = \frac{n!}{k!(n-k)!} = {n\choose k}. Q.E.D."

As I understand it says that T.n!(n-k)! = n!, but as said above, should it not be T.k!(n-k)!.

If I am wrong any kind of different explanation would be nice, since I understand the proof of taking an amount X, then taking a subamount Y, and a final subamount X without Y, here saying that the permutations in X (which is known) equals the permutations of Y and X without Y, therefore since this number most be proportional with the amount of k order subset, then one gets an equation.

Hi it's me again, since no one has answered, then I have changed this part of the text:

Thus the total possible permutations would be T.n!(n-k)! where T is the number of k-order subsets. That is because total permutations = adding n!(n-k)! the number of times equal to the number of k-order subsets = T.n!(n-k)!.

With this part:

Thus the total possible permutations would be T.k!(n-k)! where T is the number of k-order subsets. That is because total permutations = adding k!(n-k)! the number of times equal to the number of k-order subsets = T.k!(n-k)!.

If you believe I'm wrong, then please make a comment, otherwise I'll just keep editing thinking someone is sabotaging.