[Math] What role does Cauchy’s determinant identity play in combinatorics

co.combinatoricscombinatorial-identitiessymmetric-functions

When studying representation theory, special functions or various other topics one is very likely to encounter the following identity at some point:
$$\det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)}$$
This goes under the name of Cauchy's determinant identity and has various generalizations and analogous statements. There is also a lot of different proofs using either analysis or algebra. In my case I have always seen it introduced (or motivated) as an identity that plays an important role in combinatorics, but I realized that I haven't really seen this identity in a combinatorial context before. In this question I'm asking for a combinatorial interpretation of the above identity. A bonus to someone who can give such an interpretation to Borchardt's variation:
$$\det \left(\frac{1}{(x _i+y _j)^2}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)} \cdot \text{per}\left(\frac{1}{x _i +y _j}\right) _{1 \le i,j \le n}$$
(This seems a little too ambitious though, and I would be happy to accept an answer of just the first question)

Best Answer

See also pp. 397--398 of Enumerative Combinatorics, vol. 2. Cauchy's determinant is given in a slightly different but equivalent form. It is explained there that the evaluation of the determinant is equivalent to the fundamental identity $\prod(1-x_iy_j)^{-1} =\sum_\lambda s_\lambda(x)s_\lambda(y)$ in the theory of symmetric functions.

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