How to explain combinatorial identities? The setup of binomial expansion formula can be traced by two paths, one of which is "pure" proof by induction (using properties of combinatorial numbers), the other is "practical" comprehension by operation (considering subsets of a finite set).
There are some more examples of things like this.
$$\binom n 1 + 2\binom n 2 + \cdots + n\binom n n = n 2^{n - 1}$$
(Make a team out of $n$ people, and appoint a leader.)
or 
$$\binom n 0 ^2 + \binom n 1 ^2 + \cdots + \binom n n ^2 = \binom {2n} n$$
(Choose $n$ people from $n$ ladies and $n$ gentlemen.)
Sadly I cannnot figure out what this means "in real life":
$$\binom n 1 + 3\binom n 3 + \cdots = 2\binom n 2 + 4\binom n 4 + \cdots$$
Any hint will be appreciated.
(BTW: Is it always possible to "explain" combinatorial identities by "reality"? I wonder sometimes it may seem too "artificial"...)
 A: An “in real life” interpretation of the last identity is that the number of different chaired even-sized committees from $n$ people equals the number of chaired odd-sized committees from $n$ people.
Note that this is not an identity if $n=1$, as the left side is $1$ and the right side is $0$. However, it is an identity for $n>1$.
A combinatorial proof (that is, a proof not using algebra) is possible, even though it isn’t as slick as the double-counting proofs of the other identities you mention. You can prove this by describing a one-to-one correspondence, or pairing, between the committees counted on one side of the identity and the committees counted on the other side.
In other words, if you can match every even-sized chaired committee with a different odd-sized chaired committee so that all committees are matched, the number of even-sized chaired committees must equal the number of odd-sized chaired committees.
Here’s one way to do it. Since we have decided that $n>1$, we can choose two different people from the $n$ people and call them $A$ and $Z$.
Take any odd-sized chaired committee $c$. We’ll look at three disjoint and exhaustive cases.
Case 1: If $A$ is not on the committee, add $A$ as a member to get an even-sized committee. This is the match for $c$.
Case 2: If $A$ is on the committee, but is not the chair, remove $A$ from the committee (and keep the existing chair) to get the even-sized match for $c$.
Note that so far, the matches in the two cases are different, because in Case 1, the match for $c$ includes $A$ and in Case 2 it doesn’t.
The only case left to handle is if $A$ is the chair of $c$. To find the match for $c$ in this case, add or remove $Z$ from the committee, depending on whether $Z$ is a member. The matched committee has even size. The match has $A$ as chair, which did not occur in Case 1 or 2, so it is not the same as any previous match.
It remains to show that every odd-sized committee was matched (or to describe the inverse matching process). I’ll leave that up to you.
[Added: There is actually a simpler way to describe the matching, still using two named people $A$ and $Z$, that matches any chaired committee with one of different parity and that is pretty clearly a bijection, because if applied twice, it gets you the committee you started with. But I won’t spoil the fun of coming up with a description.]
