Let $\{ r_1, \dots, r_n \}$ and $\{ s_1, \dots, s_m \}$ be two sets of positive integers. If $\sum_{i = 1}^{n} r_{i} = \sum_{i = 1}^{m} s_{i}$ and $\prod_{i = 1}^{n}(r_i + 1) = \prod_{i = 1}^{m} (s_i + 1)$, does it necessarily follow that $n = m$? More generally, given the equalities, does it follow that there is a permutation $\pi \in S_{n}$ such that $\pi( \{ r_1, \dots, r_n \}) = \{ s_1, \dots, s_m \}$?
Perhaps symmetric functions play a non-trivial role here.
Update: As stated, neither question above can be answered in the affirmative (See Prof. Israel's counter-example). Let $e_k(x_1, \dots, x_n)$ denote the $k^{\text{th}}$-elementary symmetric polynomial of $n$ indeterminates.
Is there a positive integer $l < n$ such that if $e_{k}(r_1, \dots, r_n) = e_{k}(s_1, \dots, s_m)$ for $0 \leqslant k \leqslant l$ and $\prod_{i = 1}^{n}(r_i + 1) = \prod_{i = 1}^{m} (s_i + 1)$, then $n = m$ and $e_{k}(r_1, \dots, r_n) = e_{k}(s_1, \dots, s_m)$ for $0 \leqslant k \leqslant n$, or more generally, there is a permutation $\pi \in S_{n}$ such that $\pi( \{ r_1, \dots, r_n \}) = \{ s_1, \dots, s_m \}$?