a) $\sum_{i=0}^n \binom{n}{i} \binom{2n}{n-i} = \binom{3n}{n}$
b) $\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2 \binom{2n}{n} + \binom{2n}{n-1}$
c) $\sum_{i=0}^n \binom{k+i}{k} = \binom{k+n+1}{k+1}$
I am very confused and have no idea how to approach these problems? Any assistance?
On
Try to dream up descriptive committee-building stories.
In part a, you have $3n$ people: $n$ of them wear hats to work and $2n$ of them don't. You want to form any committee of size $n$ from among all $3n$ people.
The righthand side counts the number of ways to do this: gather all $3n$ people into a room and choose $n$ of them.
The lefthand side counts the same thing, but in a different way. First, choose exactly $i$ of the hat-wearing people, which can be done in $\binom{n}{i}$ ways. Now you need an additional $n-i$ people from your committee chosen from those without hats, which can be chosen in $\binom{2n}{n-i}$ ways. Summing over $i$ gives you the result.
For $a)$, it is equivalent to 1) select $n$ out of $3n$ elements, and 2) to select $i$ out of the first $n$ elements and then $n-i$ out of the remaining $2n$ elements, for some $i$. That is, each selection done according to 1) can be done according to 2), and vice versa. Thus there are equally many ways to do each, hence the equality.