Question 2 From $n$ different objects, take out $k$ to arrange into two columns, where the first column has $k_{1}$ objects, and the second column has $k_{2}\left(k_{1}+k_{2}=k, 1 \leqslant k \leqslant n\right)$ objects, how many different arrangements are there?
Now let's calculate the answer to problem 2. First, select one item from $n$ items to place at the front of the first row, which gives $n$ different choices; then select one from the remaining $n-1$ items to place at the second position of the first row, which gives $n-1$ different choices; $\cdots \cdots$ then select one from the remaining $n-k_{1}+1$ items to place at the last position of the first row, which gives $n-k_{1}+1$ different ways, and the placement of the first row is complete. Next, consider the placement of the second row, select one from the remaining $n-k_{1}$ items to place at the front of the second row, which gives $n-k_{1}$ different choices; then select one from the remaining $n-k_{1}-1$ items to place at the second position of the second row, which gives $n-k_{1}-1$ different choices; $\cdots \cdots$ finally, select one from the remaining $n-k_{1}-k_{2}+1$ items to place at the last position of the second row, which gives $n-k_{1}-k_{2}+1$ different ways. The entire placement process is now complete. By the multiplication principle, the total number of ways is
$$
\begin{aligned}
A_{n}^{k_{1}} \cdot A_{n}^{k_{2}} k_{1}= & n(n-1) \cdots\left(n-k_{1}+1\right) \\
& \cdot\left(n-k_{1}\right)\left(n-k_{1}-1\right) \cdots\left(n-k_{1}-k_{2}+1\right) \\
= & A_{n}^{k_{1}+k_{2}}=A_{n}^{k}
\end{aligned}
$$