Find a binomial coefficient equal to ${n\choose k} + 3 {n\choose k-1} + 3{n \choose k-2} + {n\choose k-3}$ 
Exercise. Find a binomial coefficient equal to:
$${n\choose k} + 3 {n\choose k-1} + 3{n \choose k-2} + {n\choose k-3}.$$

I don't really understand what we are asked to do when we are told to find a binomial coefficient equal to the sum of some combinations, I suppose that in a combinatorial way we must show that the partition of some set S can be equal to some binomial coefficient. Can someone please give a brief explanation for the exercise?
 A: Instead of doing it algebraically, you can notice that the sum can be rewritten
$$\binom30\binom{n}k+\binom31\binom{n}{k-1}+\binom32\binom{n}{k-2}+\binom33\binom{n}{k-3}\;.\tag{1}$$
Imagine that you have $3$ red balls and $n$ blue balls. Then $\binom3i\binom{n}{k-i}$ is the number of ways to choose $k$ of them so that exactly $i$ of the $k$ are red. The sum $(1)$ is therefore the number of ways to choose what from what?
A: Note that this is $$[z^k] (1+z)^3 (1+z)^n.$$
which is $$[z^k] (1+z)^{n+3} = {n+3\choose k}.$$
A: As Mann already stated in his comment, the equation
$$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$$
holds. This is really easy to see if you are familiar with the connection between the binomial coefficients and Pascal's triangle.
Try using that and $3 \cdot x = x + x + x$.
A: $$\overbrace{^{n}C_{k}+^{n}C_{k-1}}+\underbrace{^{n}C_{k-1}}+\overbrace{^{n}C_{k-1}+^{n}C_{k-2}}+\underbrace{^{n}C_{k-2}}+\overbrace{^{n}C_{k-2}+^{n}C_{k-3}}$$
Applying the identity on all the pairs specified.
$$\overbrace{^{n+1}C_{k}+^{n+1}C_{k-1}}+\underbrace{^{n+1}C_{k-1}+^{n+1}C_{k-2}}$$
I am sure you can continue from here. :)
