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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)

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It certainly comes into Greg Kuperberg's proof of the Alternating Sign Matrix conjecture: Another proof of the alternating-sign matrix conjecture, Internat. Math. Res. Notices (1996), 139-150. –  Robin Chapman Apr 7 '10 at 8:49
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up vote 16 down vote accepted

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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See Theorem 1.5 and section 2 of A Bijective Proof of Borchardt's Identity by Dan Singer.

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