Counting Problem: How to Get the Right Answer?

morphism
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Suppose you pick a k-element subset of {1, 2, ..., n}, call it A. How many of the other k-element subsets have k-1 elements in common with A?

I've been at this for quite some time, but I always overcount. Can anyone help me out? My last attempt gave me (n-k+1)k - \frac{k(k-1)}{2}, which isn't always correct. I feel I'm missing something simple.
 
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How many choices for B are there such that B is a k-element subset with k-1 elements in common with A? There are k-choices for the element of A that B doesn't have, and n-k choices for the element of AC that B does have, giving a total of k(n-k) choices for B. And if by "k-1 elements in common" you mean "at least k-1 elements in common," add 1 to include A.
 
Thanks! My problem was that I always took n-k+1 for the element of A^c that B doesn't have, which forced me to overcomplicate things thinking my reasoning was wrong, when in fact it was only a simple overcount.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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