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.
Combinatorics – Direct Proof of an Identity Regarding Symmetry of Integer Partitions
co.combinatoricsnt.number-theory
Best Answer
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, 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.