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

$r+1{}^n{C}_r+{}^n{C}_{r+1}={}^{n+1}{C}_x,$ [Given]
We have :
${}^n{C}_r+{}^n{C}_{r+1}={}^{n+1}{C}_x[\because {}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_1]\Rightarrow {}^{n+1}{C}_{r+1}={}^{n+1}{C}_x\Rightarrow r+1=x[\because {}^n{C}_r={}^n{C}_y\Rightarrow n=x+yorx=y]$