This exercise states that $C_{n,k} + C_{n,k-1} = C_{n+1,k}$, but I think that is wrong This is a problem from Degroot and Schervish's Probability and Statistics, in 1.8 section, exercise 14.
Prove that, for all positive integers $n$ and $k$ ($n\geq k$),
$C_{n,k} + C_{n,k-1} = C_{n+1,k}$. ($C$ means combination).
I cannot prove this and think that this is wrong.
Please help me.  Thank you.
 A: The exercise is correct: for any non-negative integers $n$ and $k$ with $n\geq k$, we have
$$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$$
because, if I have $n+1$ balls and one of them is red,$\newcommand{\sft}[1]{\mathsf{\text{#1}}}$
$$\begin{align}
\binom{n+1}{k}=\begin{array}{c}
\sft{# of ways of choosing}\\
k\sft{ balls from }n+1\sft{ balls}
\end{array}&=\begin{array}{c}
\sft{# of ways of choosing}\\
k\sft{ balls from }n+1\sft{ balls}\\
\sft{in which the red ball}\\
\sft{IS chosen}
\end{array}\;\;+\;\;\begin{array}{c}
\sft{# of ways of choosing}\\
k\sft{ balls from }n+1\sft{ balls}\\
\sft{in which the red ball}\\
\sft{ISN'T chosen}
\end{array}\\[0.2in]
&=\begin{array}{c}
\sft{# of ways of choosing}\\
\sft{the other }k-1\sft{ balls}\\
\sft{from the }n\sft{ non-red balls}
\end{array}+\begin{array}{c}
\sft{# of ways of choosing}\\
\sft{all }k\sft{ balls}\\
\sft{from the }n\sft{ non-red balls}
\end{array}\\[0.2in]
&=\qquad\quad\binom{n}{k-1}\quad\quad+\qquad\qquad\binom{n}{k}
\end{align}$$
