If we set $\exp(x)=\sum x^k/k!$, then $\exp(x+y)=\exp(x)\cdot \exp(y)$. In terms of coefficients it means that $(x+y)^n=\sum \frac{n!}{k!(n-k)!} x^ky^{n-k}$, i.e. just binomial expansion.


Now consider logarithm. Set $L(x):=\sum\_{k>0} x^k/k$, then $L(x)=-\log(1-x)$ in a sense, and hence $$L(u+v-uv)=L(u)+L(v),$$ i.e. $\sum (u+v-uv)^n/n=\sum (u^n+v^n)/n$, or, if we pass to coefficients of $u^av^b$ ($a,b\geq 1$), we get
$$
\sum\_k (-1)^k\frac{(a+b-k-1)!}{(a-k)!(b-k)!k!}=0
$$


The question is what is combinatorial meaning of this identity. Maybe, it is some exclusion-inclusion formula, as it is usual for alternating sums?



Combinatorial meaning of the functional equation for logarithm
As David noted, since the summands aren't in general integers, it's difficult to give a combinatorial interpretation to the formula. However, if we multiply by $a$ or $b$ we get integers and we can give a combinatorial interpretation to the identity that we obtain (though doing this destroys the symmetry between $a$ and $b$).


If we multiply by $b$, the sum may be written
$$\sum\_{k=0}^a (-1)^k \binom{a+b-k-1}{a-k}\binom{b}{k}.$$
This is a special case ($m=a+b-1$) of the more general identity
$$\sum\_{k=0}^a (-1)^k\binom{m-k}{a-k}\binom{b}{k} = \binom{m-b}{a},$$
which we can prove combinatorially. (Incidentally, this identity is a form of Vandermonde's theorem.)


To prove this formula, we start with a set $M$ of size $m$, with a subset $B$ of size $b$. To interpret $\binom{m-k}{a-k}\binom{b}{k}$, we choose a $k$-subset $K$ of $B$ and then choose an $(a-k)$-subset $C$ of $M-K$. The right side $\binom{m-b}{a}$ counts pairs $(K,C)$ in which $K$ is empty and $C$ is an $a$-subset of $M-B$. To prove the identity, we find an involution on the set of pairs $(K,C)$ not of this form that changes the parity of $|K|$. The pairs $(K,C)$ to be canceled are those in which $K\cup C$ contains at least one element of $B$. Then the involution moves the smallest element of $(K\cup C) \cap B$ from $K$ to $C$ or from $C$ to $K$.


