Question

$\left(\begin{array}{c}n\\ n-r\end{array}\right)+\left(\begin{array}{c}n\\ r+1\end{array}\right)$, whenever $0\le r\le n-1$ is equal to

• A. $\left(\begin{array}{c}n\\ r-1\end{array}\right)$
• B. $\left(\begin{array}{c}n\\ r\end{array}\right)$
• C. $\left(\begin{array}{c}n\\ r+1\end{array}\right)$
• D. $\left(\begin{array}{c}n+1\\ r+1\end{array}\right)$

Answer 1

The correct option is D $\left(\begin{array}{c}n+1\\ r+1\end{array}\right)$

$\left(\begin{array}{c}n\\ n-r\end{array}\right)+\left(\begin{array}{c}n\\ r+1\end{array}\right){=}^n{C}_{n-r}{+}^n{C}_{r+1}{\Rightarrow }^n{C}_r{+}^n{C}_{r+1}{=}^{n+1}{C}_{r+1}=\left(\begin{array}{c}n+1\\ r+1\end{array}\right)$