Question:
Prove that $\displaystyle\sum_{k = 0}^{r}\binom{n + k}{ k} = \binom{n+r+1}{r}$ whenever $n$ and $r$ are positive integers.
a.) using combinatorial argument.
b.) using Pascal's identity.
My Attempt:
a.) I've got nothing here.
b.) This one is basically algebra manipulation so I don't need help here.
I really don't have a clue where to start about the combinatorial argument. Can someone help me.
We have $n+1+r$ people lined up in a row, and want to choose $n+1$ of them. That can be done in $\binom{n+1+r}{n+1}$ ways, which is $\binom{n+1+r}{r}$, for we are rejecting $r$ people.
Let us count the number of choices another way. Suppose the first person in the line that is chosen is person $1$. Then there are $\binom{n+r}{n}$ ways to choose the others.
Maybe the first person chosen is person $2$. There are $\binom{n+r-1}{n}$ to choose the rest.
Maybe the first person chosen is person $3$. There are $\binom{n+r-2}{n}$ to choose the rest.
And so on. That gives total $$\binom{n+r}{n}+\binom{n+r-1}{n}+\binom{n+r-2}{n}+\cdots +\binom{n}{n}.$$ Recalling the combinatorial fact that $\binom{a+b}{a}=\binom{a+b}{b}$, and reversing the sum above, we get $$\binom{n}{0}+\binom{n+1}{1}+\cdots +\binom{n+r-1}{r-1}+\binom{n+r}{n},$$ which is the desired sum.
Remark: It should be easy to change the story, so that we can obtain $\binom{n+1+r}{r}$ and $\binom{n+k}{k}$ directly, and not as $\binom{n+1+r}{n+1}$ and $\binom{n+k}{n}$.
