How to prove $C(15,4) \cdot C(11,5) = C(15,5) \cdot C(10,4)$ combinatorially? 
How to prove $C(15,4) \cdot C(11,5) = C(15,5) \cdot C(10,4)$ combinatorially?

I understand how to prove this algebraically but I'm uncertain of what exactly I'm supposed to be showing to prove it combinatorially. 
By $C(n,k)$, I am referring to $n$ choose $k$, i.e., $\dbinom{n}k$.
 A: Let us prove the following combinatorially:
$$\dbinom{n}k \dbinom{n-k}r = \dbinom{n}r \dbinom{n-r}k$$
Given a set of $n$ students, we want to choose $k$ of them to be in team $1$ and $r$ of them to be in team $2$.
First let us choose the students to go to team $1$ and then to team $2$. The number of ways of doing this is $$\dbinom{n}k \dbinom{n-k}r$$
Now let us choose the students to go to team $2$ first and then to team $1$. The number of ways of doing this is $$\dbinom{n}r \dbinom{n-r}k$$
Both have to be equal. Take $n=15$, $k=5$ and $r=4$ for your case.
It also worth noting that there is another way to count the same. First choose $k+r$ students from $n$ students, and choose $k$ of them to go to team $1$ and the rest $r$ of them to go to team $2$. This can be done in $$\dbinom{n}{k+r} \dbinom{k+r}k$$
Hence, we have
$$\color{red}{\dbinom{n}k \dbinom{n-k}r} = \color{blue}{\dbinom{n}r \dbinom{n-r}k} = \color{green}{\dbinom{n}{k+r} \dbinom{k+r}k}$$
A: HINT: I have $15$ white balls, numbered $1$ through $15$. There are $\binom{15}4$ ways to pick $4$ of them and paint them blue. That leaves $11$ white balls, and there are $\binom{11}5$ ways to choose $5$ of them and paint them red. There are therefore $\binom{15}4\binom{11}5$ ways to perform the two tasks in succession, so there are $\binom{15}4\binom{11}5$ ways to paint $4$ of the $15$ balls blue and $5$ of them red.
Now suppose that you do the choosing and painting in the other order: first choose $5$ balls to paint red, then choose $4$ of the remaining white balls to paint blue. In how many ways can you do that?
Of course you’re getting the same results in each case: every possible way of painting $4$ of the balls blue and $5$ of them red. Thus, the two computations must yield the same result. Counting the same thing in two different ways like this is the essence of a combinatorial proof that two calculations yield the same result.
A: You can interpret the lhs as the number of ways to split 15 things into 4|11 and then split the 11 into 5|6, giving 4|5|6.  Similarly the rhs is the number of ways to split 15 things into 5|10 and then split the 10 into 4|6.  Therefore combinatorially both the rhs and the lhs therefore give the number of ways to split 15 things into 4|5|6.
A: As a kind of answer / amusing aside, there is a variation of this problem presented as a "self-working" magic trick in Martin Gardner's Hexaflexagons and Other Mathematical Diversions:

The magician, who is seated at a table directly opposite a spectator,
  first reverses 20 cards anywhere in the deck. That is, he turns them
  face up in the pack. The spectator thoroughly shuffles the deck so
  that these reversed cards are randomly distributed. He then holds the
  deck underneath the table, where it is out of sight of everyone, and
  counts off 20 cards from the top. This packet of 20 cards is handed
  under the table to the magician.
The magician takes the packet but continues to hold it beneath the
  table so that he cannot see the cards. "Neither you nor I," he says,
  "knows how many cards are reversed in this group of 20 that you handed
  me. However, it is likely that the number of such cards is less than
  the number of reversed cards amoung the 32 that you are holding.
  Without looking at my cards, I am going to turn a few more face-down
  cards face up and attempt to bring the number of reversed cards in my
  packet to exaclty the same number as the number of reversed cards in
  yours."
The magician fumbles with his cards for a moment, pretending that he
  can distinguish the fronts and backs of the cards by feeling them.
  Then he brings the packet into view and spreads it on the table. The
  face-up cards are counted. Their number proves to be identical with
  the number of face-up cards among the 32 held by the spectator!

I will leave it as an exercise to the reader to figure out how the trick is done, but suffice it to say it involves the same basic combinatoric problem stated here.
