【Example 1】There are 6 types of book wrapping paper, and you want to buy 10 sheets to take home for wrapping books. How many different ways are there to purchase them?
The selection process described here clearly falls under the category of combinations, as the order of the 10 selected book covers does not matter. However, there are only 6 types of book covers available, so the 10 selected book covers must include some duplicates. This is a combination problem where identical elements can be repeatedly selected, hence it is called a combination with repetition problem.

Using the occupancy problem learned earlier, it is not difficult to provide the solution to Example 1: Imagine 10 identical balls to be placed into 6 different boxes, where the number of balls in each box represents the number of corresponding book covers selected. Therefore, there are $C_{6+10-1}^{6}=C_{15}^{6}$ different ways to purchase.

Now, let's introduce a notation $\tilde{C}_{n}^{k}$, which represents the number of ways to select $k$ elements from $n$ different elements with repetition allowed. Thus, after similar considerations, we can conclude that
$$
\tilde{C}_{n}^{k}=C_{n+k-1}^{k}
$$

Using formula (1), we can conveniently calculate the number of ways for combinations with repetition.
Combinations with repetition have wide applications in people's lives, and some of the discussions are very enlightening. We will introduce some of these aspects below.

Suppose the $n$ different elements are numbered from 1 to $n$, then the $k$ selected elements correspond to a set of numbers: $i_{1}, i_{2}, \cdots, i_{k}$. If we write them in ascending order, then we have
$$
1 \leqslant i_{1} \leqslant i_{2} \leqslant \cdots \leqslant i_{k} \leqslant n
$$

It is easy to understand that the number of different combinations with repetition is the same as the number of different $k$-tuples with the above properties.
Clearly, for each such tuple, we have
$$
1 \leqslant i_{1}<i_{2}+1<\cdots<i_{k}+(k-1) \leqslant n+k-1
$$

The tuple: $i_{1}, i_{2}+1, i_{3}+2, \cdots, i_{k}+(k-1)$ is a non-repeating combination way from $\{1,2, \cdots, n+k-1\}$, and the number of different combination ways is exactly $C_{n+k-1}^{k}$. By such a correspondence, we again derive the formula $\tilde{C}_{n}^{k}=C_{n+k-1}^{k}$.
Next, let's consider another special type of combination problem.
Still consider the $k$-times unordered selection from the set $\{1,2, \cdots, n\}$. Clearly, under the usual combination method, a selected $k$-tuple (if written in ascending order) satisfies the following two inequality conditions:
(i) $1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant n$;
(ii) $i_{2}-i_{1} \geqslant 1, i_{3}-i_{2} \geqslant 1, \cdots, i_{k}-i_{k-1} \geqslant 1$.
However, in some daily life problems, the selected tuple is required to satisfy the condition for some integer $m \geqslant 2$:
$$
\begin{array}{l}
1^{0} 1 \leqslant j_{1}<j_{2}<\cdots<j_{k} \leqslant n ; \\
2^{\circ} j_{2}-j_{1} \geqslant m, j_{3}-j_{2} \geqslant m, \cdots, j_{k}-j_{k-1} \geqslant m .
\end{array}
$$

For example: In a row of 40 seats, 10 seats are reserved for children, requiring at least two adults between any two children; In a performance with 14 programs, 3 of them are dances, requiring at least 3 other programs between any two dances, etc.

The above problems can all be transformed into the problem of selecting an unordered $k$-tuple from the set $\{1,2, \cdots, n\}$, but the selected tuple should satisfy conditions $1^{\circ}$ and $2^{\circ}$, meaning they are restricted in spacing, hence they are conventionally called limited-distance combinations.

Next, we will use a similar method to derive the number of ways for limited-distance combinations. Let $\left(j_{1}, j_{2}, \cdots, j_{k}\right)$ satisfy conditions $1^{\circ}$ and $2^{\circ}$, and subtract $0, m-1, 2(m-1), \cdots, (k-1)(m-1)$ from them, respectively, and denote the resulting numbers as $j_{1}{ }^{\prime}, j_{2}{ }^{\prime}, \cdots, j_{k}{ }^{\prime}$, i.e.,
1) $1 \leqslant j_{1}{ }^{\prime}<j_{2}{ }^{\prime}<j_{3}{ }^{\prime}<\cdots<j_{k}{ }^{\prime} \leqslant n-(k-1)(m-1)$;
2) $j_{2}{ }^{\prime}-j_{1}{ }^{\prime} \geqslant 1, j_{3}{ }^{\prime}-j_{2}{ }^{\prime} \geqslant 1, \cdots, j_{k}{ }^{\prime}-j_{k-1}{ }^{\prime} \geqslant 1$.

Conversely, for any $k$ numbers $\left(j_{1}{ }^{\prime}, j_{2}{ }^{\prime}, \cdots, j_{k}{ }^{\prime}\right)$ that satisfy conditions 1) and 2), there uniquely corresponds a $k$-tuple $\left(j_{1}, j_{2}, \cdots, j_{k}\right)$ that satisfies conditions $1^{\circ}$ and $2^{\circ}$, where $j_{r}=j_{r}{ }^{\prime}+(r-1)(m-1)$. Therefore, there is a one-to-one correspondence between the two. The number of ways that satisfy conditions 1) and 2) is the number of ways to select $k$ elements from $\{1,2, \cdots, n-(k-1)(m-1)\}$, so the number of ways for limited-distance combinations is $C_{n-(k-1)(m-1)}^{k}$.