Question 908843
<pre>
{{{"nCr"}}}{{{"?=?"}}}{{{"(n-r)Cr"+2*"(n-2)C(r-1)"+"(n-2)C(r-2)"}}}

We will work with the right side:

{{{(n-2)!/(r!(n-2-r)!)}}}{{{""+""}}}{{{2*(n-2)!/(r-1)!(n^""-2-(r-1))!}}}{{{""+""}}}{{{(n-2)!/(r-2)!(n^""-2-(r-2))!}}}

{{{(n-2)!/(r!(n-2-r)!)}}}{{{""+""}}}{{{2*(n-2)!/(r-1)!(n^""-2-(r-1))!}}}{{{""+""}}}{{{(n-2)!/(r-2)!(n^""-2-(r-2))!}}}

{{{(n-2)!/(r!(n-2-r)!)}}}{{{""+""}}}{{{2*(n-2)!/(r-1)!(n-2-r+1)!}}}{{{""+""}}}{{{(n-2)!/(r-2)!(n-2-r+2)!}}}

Simplify and write letters before numbers:

{{{(n-2)!/(r!(n-r-2)!)}}}{{{""+""}}}{{{2*(n-2)!/((r-1)!(n-r-1)!)}}}{{{""+""}}}{{{(n-2)!/((r-2)!(n-r)!)}}}

Factor out (n-2)!

{{{(n-2)!( 1/(r!(n-r-2)!)+2/((r-1)!(n-r-1)!)+1/((r-2)!(n-r)!) )}}}

Write r! as r(r-1)(r-2)!
Write (r-1)! as (r-1)(r-2)!
Write (n-r)! as (n-r)(n-r-1)(n-r-2)!
Write (n-r-1)! as (n-r-1)(n-r-2)!

{{{(n-2)!(1/(r(r-1)(r-2)!(n-r-2)!)+2/((r-1)(r-2)!(n-r-1)(n-r-2)!)+1/((r-2)!(n-r)(n-r-1)(n-r-2)!   ) )}}}

The LCD is {{{r(r-1)(r-2)!(n-r)(n-r-1)(n-r-2)!}}}

{{{(n-2)!( (1red((n-r)(n-r-1)))/(LCD)+(2red(r(n-r)))/(LCD)+(1red(r(r-1)))/(LCD))}}}

(1)  {{{(n-2)!(( (n-r)(n-r-1)+2r(n-r)+r(r-1))/(LCD)) }}}

Now we simplify the numerator of (1):

{{{(n-r)(n-r-1)+2r(n-r)+r(r-1)}}}

{{{(n-r)(n-r-1)+2r(n-r)+r(r-1)}}}

We factor (n-r) out of the first two terms:

{{{(n-r)(n-r-1+2r)+r(r-1)}}}

{{{(n-r)(n+r-1)+r(r-1)}}}

{{{(n-r)((n+r)-1^"")+r(r-1)}}}

Distribute the (n-r)

{{{(n-r)(n+r)-(n-r)+r(r-1)}}}

Multiply everything out to remove all the parentheses:

{{n^2-r^2-n+r+r^2-r}}}

The r's and rē's

{{{n^2-n}}}}

{{{n(n-1)}}}

Substitute for the numerator in (1)

(1)  {{{(n-2)!(n(n-1))/(LCD)) }}}

And when we multiply (n-2)! by that numerator,

(1)  {{{(n(n-1)(n-2)!)/(LCD)) }}}

the numerator becones n!

(1)  {{{n!/(LCD)) }}}

Next we simplify the LCD

LCD = {{{r(r-1)(r-2)!(n-r)(n-r-1)(n-r-2)!}}}

That is just {{{r!(n-r)!}}}

So now (1) becomes

(1)  {{{n!/(r!(n-r)!)}}}

which is

(1)  {{{nCr}}}

Whew!

Edwin</pre>