Give a combinatorial proof: For n and k satisfying ,

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. When we factor out k(k-1)! we are left with

Now take (n-k)k(k-1)!=(n-k)k! and if we multiply by (n-k)!/(n-k)! we get (n-k)k!(n-k)!/(n-k)! = (n-k)n!/(n-k)!

This would equal except for the extra (n-k) in the numerator. Can someone please point out my mistake?