Question

Find the value of ${}^n{C}_r{+}^n{C}_{r-1}$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

we know,
${}^n{C}_r=\frac{n!}{r!(n-r)!}$
${}^n{C}_{r-1}=\frac{n!}{(r-1)!(n-r+1)!}$
Substitution the valued of ${}^n{C}_r$ and ${}^n{C}_{r+1}$ to simplify the expression
${}^n{C}_r+{}^n{C}_{r-1}=\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$
=$\frac{n!}{r\times (r-1)!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)(n-r)!}$
=$\frac{n!}{(r-1)!(n-r)!}[\frac{1}{r}+\frac{1}{n-r+1}]$
=$\frac{n!}{(r-1)!(n-r)!}\times \frac{n-r+1+r}{r(n-r+1)}=\frac{(n+1)!}{r!(n+1-r)!}={}^{n+1}{C}_r$