A little while ago in a now-deleted question, someone asked for a proof of the formula
$$\sum_{i=1}^n \binom ni i^4= 2^{n-1}n+7 \times 2^{n-2}n(n-1)+6 \times 2^{n-3}n(n-1)(n-2)+2^{n-4}n(n-1)(n-2)(n-3).$$
The usual solution is to differentiate the series for $(1+x)^n$ a few times, and that works just fine. My solution, however, was combinatorial. Even though the question has now been deleted, I thought the combinatorial approach was worth preserving. It appears below. I hope you agree.
Given an $n$-set $S$, the left-hand side is the number of ways to choose first a $k$-subset $A \subseteq S$, and then second, a length-$4$ sequence $B$ drawn from $A$ (with repetitions possible and in which order matters). So the left-hand side is the number of ordered pairs $\langle A, B \rangle$, where $A \subseteq S$ and $B \in A^4.$
Now let's look at the right-hand side. We'll see that this also counts the number of ordered pairs $\langle A, B \rangle$ as above, only this time choosing $B$ first and then counting the number of possibilities for $A$.
First choose a $4$-element sequence $B \in S^4$. How many $k$-subsets $A \subseteq S$ contain the elements of $B$? In other words, once we know $B$, how many possibilities are there for $A$? That depends on how many distinct elements $B$ has.
If $B$ has $4$ distinct elements, there are $2^{n-4}$ supersets containing $B$. There are $n(n-1)(n-2)(n-3)$ possible sequences with $4$ distinct elements. That's because there are $\binom n4$ ways to choose $4$ distinct elements and $24$ ways to order them.
If $B$ has $3$ distinct elements, there are $2^{n-3}$ supersets containing $B$. There are $6n(n-1)(n-2)$ possible sequences with $3$ distinct elements. That's because there are $\binom n3$ ways to choose the $3$ distinct elements, $3$ ways to choose which of them is a duplicate, and $12$ ways to order the sequence once its elements and the duplicate are known.
If $B$ has $2$ distinct elements, there are $2^{n-2}$ supersets containing $B$. There are $7n(n-1)$ possible sequences with $2$ distinct elements. That's because there are $\binom n2$ ways to choose the two distinct elements, with $8$ ways to order them if they're distributed $3$-$1$, and $6$ ways to order them if they're distributed $2$-$2$.
Finally, if $B$ has only $1$ distinct element, there are $2^{n-1}$ supersets containing $B$. There are $n$ possible sequences with $1$ distinct element.
Adding up these possibilities gives the right-hand side.