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.
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,$\newcommand{\sft}[1]{\mathsf{\text{#1}}}$ $$\begin{align} \binom{n+1}{k}=\begin{array}{c} \sft{# of ways of choosing}\\ k\sft{ balls from }n+1\sft{ balls} \end{array}&=\begin{array}{c} \sft{# of ways of choosing}\\ k\sft{ balls from }n+1\sft{ balls}\\ \sft{in which the red ball}\\ \sft{IS chosen} \end{array}\;\;+\;\;\begin{array}{c} \sft{# of ways of choosing}\\ k\sft{ balls from }n+1\sft{ balls}\\ \sft{in which the red ball}\\ \sft{ISN'T chosen} \end{array}\\[0.2in] &=\begin{array}{c} \sft{# of ways of choosing}\\ \sft{the other }k-1\sft{ balls}\\ \sft{from the }n\sft{ non-red balls} \end{array}+\begin{array}{c} \sft{# of ways of choosing}\\ \sft{all }k\sft{ balls}\\ \sft{from the }n\sft{ non-red balls} \end{array}\\[0.2in] &=\qquad\quad\binom{n}{k-1}\quad\quad+\qquad\qquad\binom{n}{k} \end{align}$$
The other answer had a combinatorial proof of the identity, just in case you still need it, here I will present a computational proof of the same. $$\binom{n}{k}+\binom{n}{k-1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}=n!\left[\frac{1}{k(k-1)!(n-k)!}+\frac{1}{(k-1)!(n-k+1)(n-k)!}\right]$$ $$=\frac{n!}{(k-1)!(n-k)!}\left[\frac{1}{k}+\frac{1}{n-k+1}\right]$$ $$=\frac{n!}{(k-1)!(n-k)!}\left[\frac{n+1}{k(n-k+1)}\right]$$ $$=\frac{(n+1)n!}{k(k-1)!\cdot (n-k+1)(n-k)!}$$ $$=\frac{(n+1)!}{k!(n+1-k)!}$$ $$=\binom{n+1}{k}$$ as required.
