Q.Using the factorial formula $^{n}C_r = \dfrac{n!}{r!(n-r)!}$, prove algebraically that $^{n}C_{r-1} + {}^{n}C_r = {}^{n+1}C_r$ for $1 \le r \le n$.

[!TLDR] Write both combinations using $n!$ factorial form, combine over a common denominator, and simplify. [!ANSWER] $^{n}C_{r-1}+{}^{n}C …

Using $^{n}C_r=\dfrac{n!}{r!(n-r)!}$: $^{n}C_{r-1}+{}^{n}C_r=\dfrac{n!}{(r-1)!(n-r+1)!}+\dfrac{n!}{r!(n-r)!}=\dfrac{n!}{(r-1)!(n-r)!}\left[\dfrac{1}{n-r+1}+\dfrac{1}{r}\right]$. Combining the bracket over a common denominator: $\dfrac{1}{n-r+1}+\dfrac{1}{r}=\dfrac{r+(n-r+1)}{r(n-r+1)}=\dfrac{n+1}{r(n-r+1)}$. So $^{n}C_{r-1}+{}^{n}C_r=\dfrac{n!}{(r-1)!(n-r)! …

Write both combinations in factorial form, factor out the common part $\dfrac{n!}{(r-1)!(n-r)!}$, combine the remaining fractions over a common denomina …

Errors combining $(r-1)!(n-r+1)!$ and $r!(n-r)!$ into a common factored form are common; it helps to note $(n-r+1)!=(n-r+1 …