Question 10: Let $\mathrm{A}$ and $\mathrm{B}$ be non-empty sets, and $\mathrm{A} \cap \mathrm{B}=\Phi, \mathrm{A} \cup \mathrm{B}=\{1,2, \ldots, \mathrm{n}\}(\mathrm{n} \geq 3, \mathrm{n} \in$ $\left.N^{+}\right)$, denote the number of elements in A and B as $a$ and $b$, respectively. The number of all ordered set pairs $(A, B)$ that satisfy “$a \in B$, and $b \in A$” is $\mathrm{a}_{\mathrm{n}}$. Find the expression for $\mathrm{a}_{\mathrm{n}}$.
When $n=2 k+1 (k \in N^{+})$, if $|A|=a$, then $|B|=2 k+1-a$. Clearly, $1 \leq a \leq 2 k$, and $a \neq 2 k+1-a$. Thus, $a \in B$ and $2 k+1-a \in A$. At this point, from the remaining $2 k-1$ numbers, any $a-1$ numbers can be chosen to be placed in set $A$, and the remaining numbers can be placed in set $B$, all satisfying the conditions. Therefore, $a_{2 k+1}=\sum_{a=1}^{2 k} C_{2 k-1}^{a-1}=\sum_{a=0}^{2 k-1} C_{2 k-1}^{a}=2^{2 k-1}$.

When $n=2 k+2 (k \in N^{+})$, if $|A|=a$, then $|B|=2 k+2-a$. Clearly, $1 \leq a \leq 2 k+1$, and $a \neq k+1$. Thus, $a \in B$ and $2 k+2-a \in A$. At this point, from the remaining $2 k$ numbers, any $a-1$ numbers can be chosen to be placed in set $A$, and the remaining numbers can be placed in set $B$, all satisfying the conditions. Therefore, $a_{2 k+2}=\sum_{a=1}^{2 k+1} C_{2 k}^{a-1}-C_{2 k}^{k}=\sum_{a=0}^{2 k} C_{2 k}^{a}-C_{2 k}^{k}=2^{2 k}-C_{2 k}^{k}$.