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Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i<j\leq m}(x_i-x_j).$$ The number of plane partitions in an $n\times m\times m$ box (MacMahon) is given by the neat product formula $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$

Notation. Let $\mathcal{F}_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j_1,\dots,j_m\}$ for $\mathbf{J}\in\mathcal{F}_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.

I would like to ask:

QUESTION. Is this expansion true? A combinatorial proof is desired, if possible. $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2} =\sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$

Remark. Observe the cute fact $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer (let's make Fedor's comment explicit).

Proof. Replacing $\mathbf{x}=(x_1,\dots,x_m)$ by $\mathbf{J}=(j_1,\dots,j_m)$ into $$\det\left[\binom{x_j}i\right] =\frac1{\prod_{i<j}(j-i)}\det\left[\prod_{k=0}^{i-1}(x_j-k)\right] =\frac{\det\left[x_j^{i-1}\right]}{\prod_{i<j}(j-i)} =\prod_{1\leq i \lt j\leq m} \frac{x_j-x_i}{j-i}.$$ results in integer entries, and thus integer determinant. $\qquad\square$

Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i<j\leq m}(x_i-x_j).$$ The number of plane partitions in an $n\times m\times m$ box (MacMahon) is given by the neat product formula $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$

Notation. Let $\mathcal{F}_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j_1,\dots,j_m\}$ for $\mathbf{J}\in\mathcal{F}_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.

I would like to ask:

QUESTION. Is this expansion true? A combinatorial proof is desired, if possible. $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2} =\sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$

Remark. Observe the cute fact $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer (let's make Fedor's comment explicit).

Proof. Replacing $\mathbf{x}=(x_1,\dots,x_m)$ by $\mathbf{J}=(j_1,\dots,j_m)$ into $$\det\left[\binom{x_j}i\right] =\frac1{\prod_{i<j}(j-i)}\det\left[\prod_{k=0}^{i-1}(x_j-k)\right] =\frac{\det\left[x_j^{i-1}\right]}{\prod_{i<j}(j-i)} =\prod_{1\leq i \lt j\leq m} \frac{x_j-x_i}{j-i}.$$ results in integer entries, and thus integer determinant $\frac{V(\mathbf{J})}{V(\mathbf{I})}$. $\qquad\square$

Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i<j\leq m}(x_i-x_j).$$ The number of plane partitions in an $n\times m\times m$ box (MacMahon) is given by the neat product formula $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$

Notation. Let $\mathcal{F}_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j_1,\dots,j_m\}$ for $\mathbf{J}\in\mathcal{F}_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.

I would like to ask:

QUESTION. Is this expansion true? A combinatorial proof is desired, if possible. $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2} =\sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$

Remark. It is a cute observation that $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer.

Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i<j\leq m}(x_i-x_j).$$ The number of plane partitions in an $n\times m\times m$ box (MacMahon) is given by the neat product formula $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$

Notation. Let $\mathcal{F}_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j_1,\dots,j_m\}$ for $\mathbf{J}\in\mathcal{F}_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.

I would like to ask:

QUESTION. Is this expansion true? A combinatorial proof is desired, if possible. $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2} =\sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$

Remark. Observe the cute fact $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer (let's make Fedor's comment explicit).

Proof. Replacing $\mathbf{x}=(x_1,\dots,x_m)$ by $\mathbf{J}=(j_1,\dots,j_m)$ into $$\det\left[\binom{x_j}i\right] =\frac1{\prod_{i<j}(j-i)}\det\left[\prod_{k=0}^{i-1}(x_j-k)\right] =\frac{\det\left[x_j^{i-1}\right]}{\prod_{i<j}(j-i)} =\prod_{1\leq i \lt j\leq m} \frac{x_j-x_i}{j-i}.$$ results in integer entries, and thus integer determinant. $\qquad\square$

Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i<j\leq m}(x_i-x_j).$$ The number of plane partitions in an $n\times m\times m$ box (MacMahon) is given by the neat product formula $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$

Notation. Let $\mathcal{F}_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j_1,\dots,j_m\}$ for $\mathbf{J}\in\mathcal{F}_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.

I would like to ask:

QUESTION. Is this expansion true? A combinatorial proof is desired, if possible. $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2} =\sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$

Consider the Vandermonde's determinant computed by $$V(x_1,\dots,x_m):=\det(x_j^{i-1})_{i,j=1}^m=\prod_{1\leq i<j\leq m}(x_i-x_j).$$ The number of plane partitions in an $n\times m\times m$ box (MacMahon) is given by the neat product formula $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$

Notation. Let $\mathcal{F}_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j_1,\dots,j_m\}$ for $\mathbf{J}\in\mathcal{F}_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.

I would like to ask:

QUESTION. Is this expansion true? A combinatorial proof is desired, if possible. $$\prod_{i=1}^n\prod_{j=1}^m\prod_{k=1}^m\frac{i+j+k-1}{i+j+k-2} =\sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$

Remark. It is a cute observation that $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer.