Question

Factorial and algebraic fraction question:

Answer 1

I know it seems easy but I just don't know how to get from the one line to the next (indicated by red arrow in picture attached)

Answer 2

I understand the rest of it and i understand how to get to the first line

Answer 3

@Hero @TuringTest @estudier

Answer 4

so the deal is to get a common denominator, which is k!(n-k+2)!
the first term is over k!(n-k)!
since (n-k+2) is two more than (n-k) if we do\[\frac{(n-k+2)!}{(n-k)!}\]that will leave two highest terms to multiply together in the denominator, so\[\frac{(n-k+2)!}{(n-k)!}=(n-k+2)(n-k+1)\]hence that is left on the numerator when the first term is put over a common denominator
the same deal with the rest, do you see it now?

Answer 5

one way to see it:\[\frac{(n-k+2)!}{(n-k)!}=\frac{(n-k+2)(n-k+1)\cancel{(n-k)!}}{\cancel{(n-k)!}}\]\[=(n-k+2)(n-k+1)\]

Answer 6

so\[\frac{n!}{k!(n-k)!}\cdot\frac{(n-k+2)!}{(n-k+2)!}=\frac{n!(n-k+2)(n-k+1)}{k!(n-k+2)!}\]\[\frac{2n!}{(k-1)!(n-k+1)!}\cdot\frac{k!(n-k+2)!}{k!(n-k+2)!}=\frac{2n!k(n-k+2)}{k!(n-k+2)!}\]\[\frac{n!}{(k-2)!(n-k+2)!}\cdot\frac{k!}{k!}=\frac{n!k(k-1)}{k!(n-k+2)!}\]

Answer 7

deal this with