Consider selecting \(k\) from \(n\) with replacement but ignoring order.

1. Let’s take the specific case of \(k=2\) and \(n=4\). List out all possible ways you can select \(k=2\) items from \(\{1, 2, 3, 4\}\) in such a scenario.
2. Let \(x_i\) be the number of times that element \(\{i\}\) is selected, for each \(i = 1,\,2,\,3,\, 4\). For each combination identified in (a), show that it satisfies the following equation:

\[x_1 + x_2 + x_3 + x_4 = 2, \quad x_1, x_2, x_3, x_4 \in \{0, 1, 2\}.\]

3. State an analogous equation to part b for general \(n\) and \(k\). Use this to show that the number of ways you can select \(k\) from \(n\) with replacement and ignoring order is

\[ \binom{n+k-1}{k}=\frac{(n+k-1)!}{k!(n-1)!}. \]