Suppose $(a\_1,\ldots, a\_k)$ is an integer partition of $n$, and $(b\_1,\ldots,b\_k)$ is a rearrangement of the $a$-sequence. Prove the following identity (preferably combinatorially):
$$
\sum\_{j\_2+\cdots+j\_k=l,\atop l\geq 0, \, a\_t>j\_t\geq 0} \quad\frac{(-1)^l l!}{(a\_1+l+1)\_{l+1}} {a\_2\choose j\_2}\cdots {a\_k \choose j\_k}=\sum\_{j\_2+\cdots+j\_k=l,\atop l\geq 0, \, b\_t>j\_t\geq 0} \quad\frac{(-1)^l l!}{(b\_1+l+1)\_{l+1}} {b\_2\choose j\_2}\cdots {b\_k \choose j\_k},
$$
where $(x)\_k$ is the falling factorial.



direct proof of an identity regarding certain symmetry of integer partitions?
I assume you meant $j\_t\leq a\_t$ (not $j\_t<a\_t$).


It's sufficient to prove that for any analytic function $f(x)$, the function
$$F(a\_1,a\_2) := \sum\_{l\geq 0} \sum\_{j=0}^{a\_2} \frac{(-1)^ll!}{(a\_1+l+1)\_{l+1}} \binom{a\_2}{j} [x^{l-j}]\ f(x)$$
is symmetric, i.e. $F(a\_1,a\_2) = F(a\_2,a\_1)$.


Using the property of [beta function](https://en.wikipedia.org/wiki/Beta_function), we have


\begin{split}
F(a\_1,a\_2) &=\sum\_{l\geq 0} \frac{(-1)^ll!}{(a\_1+l+1)\_{l+1}} [x^l]\ (1+x)^{a\_2}f(x) \\
&=\sum\_{l\geq 0} (-1)^l \int\_0^1 t^l (1-t)^{a\_1} {\rm d}t\ [x^l]\ (1+x)^{a\_2}f(x)\\
&=\int\_0^1 (1-t)^{a\_1} (1-t)^{a\_2} f(-t) {\rm d}t,
\end{split}
which is clearly symmetric.


