11. (POL 1) Find the number of partitions of the set $\{1,2, \ldots, n\}$ into three subsets $A_{1}, A_{2}, A_{3}$, some of which may be empty, such that the following conditions are satisfied: (i) After the elements of every subset have been put in ascending order, every two consecutive elements of any subset have different parity. (ii) If $A_{1}, A_{2}, A_{3}$ are all nonempty, then in exactly one of them the minimal number is even.
11. Let \( A_{1} \) be the set that contains 1, and let the minimal element of \( A_{2} \) be less than that of \( A_{3} \). We shall construct the partitions with required properties by allocating successively numbers to the subsets that always obey the rules. The number 1 must go to \( A_{1} \); we show that for every subsequent number we have exactly two possibilities. Actually, while \( A_{2} \) and \( A_{3} \) are both empty, every successive number can enter either \( A_{1} \) or \( A_{2} \). Further, when \( A_{2} \) is no longer empty, we use induction on the number to be placed, denote it by \( m \): if \( m \) can enter \( A_{i} \) or \( A_{j} \) but not \( A_{k} \), and it enters \( A_{i} \), then \( m+1 \) can be placed in \( A_{i} \) or \( A_{k} \), but not in \( A_{j} \). The induction step is finished. This immediately gives us that the final answer is \( 2^{n-1} \).