Question

Find ${}^n{C}_r{+}^n{C}_{r-1}=.....$ for $n=r=3$

Answer 1

${{}^{n}C}_r+{{}^{n}C}_{r-1}=?$

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

${{}^{n}C}_{r-1}=\frac{n!}{(r-1)!(n-r+11)!}$

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

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

$=\frac{n!}{(r-1)!(n-r+1)!}\left[\frac{n-r+1+r}{r(n-r+1)}\right]$

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

$={{}^{n+1}C}_r$

putting values we get $={{}^{3+1}C}_3=4$