Question

If $1\le r\le n$, then prove that ${{}^{n}C}_r+{{}^{n}C}_{r-1}={{}^{n+1}C}_r$

Answer 1

${}^n{C}_r{+}^n{C}_{r-1}=\frac{n!}{(n-r)!r!}+\frac{n!}{(n-r+1)!(r-1)!}$

$=\frac{n!}{(n-r+1)!r!}(n-r+1+r)$, factor out common factor

$=\frac{n!}{(n-r+1)!r!}(n+1)$

$=\frac{(n+1)!}{\left[\right(n+1)-r]!r!}{=}^{n+1}{C}_r$, Hence proved

Fact: ${}^n{C}_r=\frac{n!}{(n-r)!r!}$