0

enter image description here

At the second line we get two fractions with unlike denominators. To add them, I thought you'd need to get them to the common denominator (n-k)!k!(n-k+1)!(k-1)!

But instead, the proof does three things that confuse me:

1) the green fraction is only multiplied by (n-k+1)/(n-k+1). I expected ((n-k+1)!(k-1)!)/((n-k+1)!(k-1)!)

2) the yellow fraction is only multiplied by k/k. I expected ((n-k)!k!)/((n-k)!k!)

3) After multiplying those green and yellow fractions by those terms, you are still left with two fractions with unlike denominators [unless I'm not seeing how to simplify]. I don't see how he winds up with the expression in the purple box.

Ugh. I'm probably making some very obvious Algebra mistake, but I cannot see it.

(this proof came from a different stackoverflow question I was interested in)

Community
  • 1
tuple_linear
  • 283

1 Answers1

1

The factorials have lots of common terms. $(k-1)!$ is the product of all the numbers from $k-1$ down to $1$ while $k!$ is the product of all the numbers from $k$ down to $1$, so we have $k!=k(k-1)!$. Similarly we have $(n-k+1)!=(n-k+1)(n-k)!$ That is why they multiplied top and bottom by $n-k+1$ in the first term and $k$ in the second going from the second line to the third. From the third to the fourth they used the earlier relations, which get the two fractions over a common denominator.

Ross Millikan
  • 374,822
  • yeah, I totally missed the k! = k(k-1)! strategy. I don't know how to become more adept at this sort of stuff. It's very frustrating when I face proof exercises. – tuple_linear Nov 28 '16 at 15:03