I found a way. As an exit slip of ORE, here is my way. First, let function f map a factorial to its number, like f(120) = 5. If I use normal math terms it'd be f: x! → x. Now, the binomial coeffecient of n and k is n!/k!(n-k)!. This can be represented differently, but let's use this definition because it has (n-k)!. In order to remove that ugly n! numerator we multiply the coeffecient by 1/n!. To remove the k! at the bottom, we multiply everything by k!. Now we have k!/n! * binom(n,k) = 1/(n-k)!, so let's do 1/binom(n,k) to get (n-k)!. Now we have n!/k! * 1/binom(n,k) = (n-k)!. To simplify this, we can remember reciprocals from pre-algebra (did that two months ago, fuck prealgebra) and we know a * 1/b = a/b, so we'd get (n!/k!)/binom(n,k) = (n-k)!. Now, remember our function f? To map (n-k)! without using subtraction, we do f((n!/k!)/binom(n,k)) = n-k. Now to transform n-k to n+k? Simple; take the conjugate of this. Writing this in equation form makes:
Well I'm lazy, go onto http://www.codecogs.com/latex/eqneditor.php and type in f\overline{\left (\frac{n!/k!}{\binom{n}{k}}\right )}=n+k in the thing. The line over everything is the conjugate symbol. Take that aft!
Fuck... nobody will understand my post I just realized.
Well I'm lazy, go onto http://www.codecogs.com/latex/eqneditor.php and type in f\overline{\left (\frac{n!/k!}{\binom{n}{k}}\right )}=n+k in the thing. The line over everything is the conjugate symbol. Take that aft!
Fuck... nobody will understand my post I just realized.