TITLE: What's the motivation of entropy as a combinatorical tool? What problems is it able to solve?
QUESTION [32 upvotes]: I am interested in using Shannon's entropy in combinatorics. It is often presented with a motivation of how much information can be passed, but assume I am not interested in that, I want to understand why it is useful as a tool to bound objects (with examples now).
Several examples that demonstrate that very nontrivial results follow from relatively entropy arguments are:

Bregman's inequality about permanents.
Whiteny's inequality about volume and its projections.
Shearer's lemma and numerous consequences; bounds on the number of independent sets in bipartite d-regular graph, number of homomorphisms to graph...
The largest size of an intersection unique family (see https://link.springer.com/article/10.1007/BF02579460 ).

If you could help me with the following I'd be very happy;

How would someone just interested in counting and bounding combinatorics arrive at entropy? 
A common theme (noted by my proffessor when teaching about it) to some of the results are that original proofs used convexity arguments, how healthy is it to be think of entropy in combinatorics as a clean way to state convexity arguments and deriving nontrivial results by assigning weights to objects and bounding sums?
When can proofs be translated to entropy proofs? For instance I find it strange there is the $k-intersection$ proof with entropy, while I'm not aware of an entropy proof for sperner's theorem about antichains.
When are problems suspectible to entropy methods?

Cheers.

REPLY [3 votes]: I wanted to share another answer, that I feel comfortable with after trying to look for examples and intuition.
We start with the following riddle:
Find the least $k(n)$ so that there are $k$ subsets $D_1,...,D_k$ of $[n]$ s.t $|D_1\cap A|,...,|D_k \cap A|$ uniquely determine $A$ for each $A\subset[n]$.
This is an interesting question (coming from- there are $n$ bits, at each turn you pick a subset and are told how many of them are turned on, but you ask the questions nonadaptively, and need to determine the bits).
An easy initial observation is this;
Each intersections has at most $n+1$ options, and so $(n+1)^k\geq 2^n$, so $k\geq n/log_2(n)$.
Another way to phrase this, is that we can encode $n$ bits using $k\cdot log_2(n+1)$. Entropy for me, is all about encoding things. Now here will come the twist of entropy, it is not about deterministically encoding things, but rather, encoding most of the things, that is, we want to be able to encode\decode a random message of the possible inputs (with the given distribution on it). This is formalized, and this is where the expression for entropy most naturally arises for me, in the theorem of Shannon:
https://en.wikipedia.org/wiki/Shannon%27s_source_coding_theorem
I won't repeat it here, but the point is that we have $X_1,..,X_n$ iid in a probability space $W$. Given $w$ in our world, we define the random variable $Y(w)$ to be $P(w' s.t X_1(w')=X_1(w),X_2(w')=X_2(w)..)$. We can calculate the expectation, but we don't have any theorem to handle concentration of product. So we do the usual, we take $log$. From this and the large law of numbers we easily obtain the theorem. 
It is nice because some easy corollaries are: $H(X,Y)\leq H(X)+H(y)$, $H(uniform)$ is maximizes given size of range.
Back to the riddle, the point is that if we take a bunch copies of random $A$, if we insisted to deterministically encode each one, we'd get the same bound. 
But there are $2$ excellent twists: 

For uniform distribution, even though we now allow to encode only with h.p, it is still hard and demands many bits.
For other distributions, this "tensorization" made a difference, a formula we can calculate.

Indeed, we have $n=H(uniform A)=H(|D_1\cap A|,...,|D_k \cap A|)\leq \sum H(|D_i \cap A|)$. So we want to bound $H(|D_i \cap A|)$.
Now if we're brute, and just insist on deterministically encoding all possible values of $|D_i \cap A|$, we don't get any improvement, but the whole point is that we allow us to only do so with h.p.
Since we expect this thing to be concentrated around $|D_i|/2$ with an interval of around $~\sqrt(|D_i|)$ around, we need only ~$log(\sqrt|D_i|$ bits, which is about half our original! (i.e we got a bound of about $2n/log(n)$).
To summarize, the trick is to tensorize (maybe there is a slight connection to the proof of the crossing inequality) by allow a small degree of freemdom, which traps uniform the uniform distribution.