I'm stuck on a simplification, used to prove $C(n – 1, r – 1) + C(n – 1, r) = C(n, r)$
Could somebody clarify the step(s) from: $\frac{(n – 1)!}{(r – 1)!(n – r)!} + \frac{(n – 1)!}{r!(n – r – 1)!}$ to $\frac{r(n – 1)!}{r!(n – r)!} + \frac{(n – r)(n – 1)!}{r!(n – r)!}$ (or $\frac{r(n – 1)! +(n – r)(n – 1)!}{r!(n – r)!}$).
I'm stuck making the denominators common via multiplication: I can't seem to work away the factorial expressions. $\frac{(n-1)!r!(n-r-1)!}{(r-1)!(n-r)!r!(r-n-1)!} + \frac{(n-1)!(r-1)!(n-r)!}{r!(n-r-1)!(r-1)!(n-r)!}$.
Any help would be very welcome!
Best Answer
Since $$ r!=r(r-1)! $$ and $$ (n-r)!=(n-r)(n-r-1)!, $$ we have the equivalent algebraic identities: $$\begin{eqnarray*} (r-1)! &=&\frac{r!}{r} \\ (n-r-1)! &=&\frac{(n-r)!}{n-r}. \end{eqnarray*}$$ So \begin{eqnarray*} \frac{(n-1)!}{(r-1)!(n-r)!}+\frac{(n-1)!}{r!(n-r-1)!} &=&\frac{(n-1)!}{\frac{ r!}{r}(n-r)!}+\frac{(n-1)!}{r!\frac{(n-r)!}{n-r}} \\ &=&\frac{r(n-1)!}{r!(n-r)!}+\frac{(n-r)(n-1)!}{r!(n-r)!} \\ &=&\frac{r(n-1)!+(n-r)(n-1)!}{r!(n-r)!}\\ &=&\frac{(r+n-r)(n-1)!}{r!(n-r)!} \\ &=&\frac{n(n-1)!}{r!(n-r)!} \\ &=&\frac{n!}{r!(n-r)!}. \end{eqnarray*}