I want the combinatoric proof of the following identity:
\begin{align*}
{\binom{j-1}{m-1}=\sum_{k=m}^j(-1)^{k-m}\binom{j}{k}}
\end{align*}
i.e. the left hand side choose what $(m-1)$ out of what $(j-1)$, I need a concrete example.
Note after solved this: Just use the Pascal Rule, since the combinatorics proof can't explain why the result is always positive.
Best Answer
Say we have $j$ boys and a coach want to form a group of at least $m$ members.
The coach claim that he can form as many groups with even number of boys as groups with odd number of boys. His reasoning: for any even group, he can make the oldest boy of these $j$ join or leave the group to make it an odd group and vice versa.
The assistant coach disagree. His reasoning: there are groups with exactly $m$ boys including the oldest. The coach cannot make the oldest boy leave the group because then the group will have less than $m$ members. There are $\binom{j-1}{m-1}$ groups like this.
$$ \rule{6cm}{0.4pt} $$
Here is an alternative for non combinatorial proof.
We use identity $\binom{n}{i}=\binom{n-1}{i-1}+\binom{n-1}{i}$
$$ \begin{align} \sum_{k=m}^{j}{\left(-1\right)^{k—m}\binom{j}{k}}&=\binom{j}{m}+\binom{j}{m+2}+...-\binom{j}{m+1}-\binom{j}{m+3}-...\\ &=\binom{j-1}{m-1}+\binom{j-1}{m}+\binom{j-1}{m+1}+\binom{j}{m+2}+...-\binom{j-1}{m}-\binom{j-1}{m+1}-\binom{j-1}{m+2}-\binom{j-1}{m+3}-...\\ &=\binom{j-1}{m-1} \end{align} $$