I'm trying to show that in Pascal's triangle, a number is the the sum of the two terms in the row above it. My working is as follows:
$\frac{n!}{(r-1)!(n-(r-1))!}$+$\frac{n!}{r!(n-r)!}$
= $\frac{n!}{(r-1)!(n-r+1) \times (n-r)!}$+$\frac{n!}{r \times (r-1)!(n-r)!}$
= $\frac{r\times(r-1)!(n-r)!n!+(r-1)!(n-r+1)\times(n-r)!n!}{(r-1)!(n-r+1)\times(n-r)!r\times(r-1)!(n-r)!}$
= $\frac{r\times(n-r)!n!+(n-r+1)\times(n-r)!}{(n-r+1\times(n-r)!r(n-r)!n!}$
= $\frac{n!\times r + n! \times (n-r+1)}{r(n-r+1)!}$,
which seems to be missing an ${r!}$ as per the fourth line of the given solution.
The textbook solution is given as follows:
${n\choose r-1}$$x^{r-1}$ + ${n\choose r}$$x^{r}$
= $\frac{n!}{(r-1)!(n-(r-1))!}$+$\frac{n!}{r!(n-r)!}$
= $\frac{n!}{(r-1)!(n-r+1) \times (n-r)!}$+$\frac{n!}{r \times (r-1)!(n-r)!}$
= $\frac{n!\times r+n!\times(n-r+1)}{r!(n-r+1)!}$
= $\frac{n!(r+n-r+1)}{r!(n-r+1)!}$
= $\frac{(n+1)!}{r!(n+1-r)!}$
= ${n+1\choose r}$$x^{r}$
Assuming the given solution is correct, I would like some guidance on where I have gone wrong.