Question

If ${}^n{C}_r{+}^n{C}_{r+1}{=}^{(n+1)}{C}_x$, then $x=$

• A. $r$
• B. $r-1$
• C. $n$
• D. $r+1$

Answer 1

The correct option is D $r+1$

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

$\therefore x=r+1$