For a non-empty integer set $A$, if it satisfies
$$
a \in A, a-1 \notin A, a+1 \notin A,
$$

then $a$ is called an isolated element of set $A$. Question: For the set
$$
M=\{1,2, \cdots, n\}(n \geqslant 3)
$$

how many $k(k \geqslant 3)$-element subsets of $M$ have no isolated elements?
Let the number of $k$-element subsets of set $M$ without isolated elements be denoted as $P_{k}$.

Let $r=\min \left\{\left[\frac{k}{2}\right], n-k+1\right\}([x]$ represents the greatest integer not exceeding the real number $x$). Then,
$$
P_{k}=\sum_{l=1}^{r} \mathrm{C}_{k-l-1}^{l-1} \mathrm{C}_{n-k+1}^{l} \text {. }
$$

First, we prove a lemma.
Lemma Let $k, l \in \mathbf{Z}_{+}$, and $k \geqslant 3, l \leqslant\left[\frac{k}{2}\right]$, $k_{i} \in \mathbf{Z}_{+}(i=1,2, \cdots, l)$ and $k_{i} \geqslant 2$. Then the number of ordered tuples $\left(k_{1}, k_{2}, \cdots, k_{l}\right)$ that satisfy the indeterminate equation $k_{1}+k_{2}+\cdots+k_{l}=k$ is $\mathrm{C}_{k-l-1}^{l-1}$.
Proof Let $k_{i}^{\prime}=k_{i}-1$.
Then $k_{i}^{\prime} \geqslant 1$.
The indeterminate equation in the problem becomes
$k_{1}^{\prime}+k_{2}^{\prime}+\cdots+k_{l}^{\prime}=k-l$.
And $l \leqslant\left[\frac{k}{2}\right] \Rightarrow l \leqslant \frac{k}{2}$
$\Rightarrow k-l-1 \geqslant l-1$.
Using the stars and bars method, we know that the number of solutions to equation (1), i.e., the number of ordered tuples $\left(k_{1}^{\prime}, k_{2}^{\prime}, \cdots, k_{l}^{\prime}\right)$, is $\mathrm{C}_{k-l-1}^{l-1}$.

Thus, the number of ordered tuples $\left(k_{1}, k_{2}, \cdots, k_{l}\right)$ that satisfy the original indeterminate equation is $\mathrm{C}_{k-l-1}^{l-1}$.
Returning to the original problem.
Let the $k$-element subset of set $M$ be $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$, where, $a_{1}a_{k_{1}}+1$, similar to the process of obtaining equation (2), we know
$a_{k_{1}+2}=a_{k_{1}+1}+1$.
Thus, we can assume that $a_{k_{1}+1}, a_{k_{1}+2}, \cdots, a_{k_{1}+k_{2}}\left(k_{2} \geqslant 2\right)$ are $k_{2}$ consecutive positive integers.

Proceeding similarly, the $k(k \geqslant 3)$-element subset of set $M$ without isolated elements, when arranged in ascending order, must consist of one or two or more segments of consecutive positive integers, each segment having a length of at least 2.

Assume there are $l(l \geqslant 1)$ segments, with the number of consecutive positive integers in each segment being $k_{i}(i=1,2, \cdots, l)$, and
$k_{1}+k_{2}+\cdots+k_{l}=k$, where, $k_{i} \in \mathbf{Z}_{+}, k_{i} \geqslant 2, i=1,2, \cdots, l$.

The above equation represents the partition of $k$ into $l$ ordered positive integers, each at least 2.

By the lemma, we know that the number of ordered tuples $\left(k_{1}, k_{2}, \cdots, k_{l}\right)$ (the number of partition methods) is $\mathrm{C}_{k-l-1}^{l-1}$.

For a given ordered partition $\left(k_{1}^{*}, k_{2}^{*}, \cdots, k_{l}^{*}\right)$, i.e., there are $l$ segments, with the number of consecutive positive integers in each segment being $k_{1}^{*}, k_{2}^{*}, \cdots, k_{l}^{*}$, the number of $k$-element subsets without isolated elements that satisfy this condition is $\mathrm{C}_{n-k+1}^{l}$.
Below is the proof.
Assume there are $n$ identical balls (not numbered yet).

【Step 1】Take $k$ balls from these $n$ balls and divide them into $l$ groups, with the number of balls in each group being $k_{1}^{*}, k_{2}^{*}, \cdots, k_{l}^{*}$;

【Step 2】Arrange the remaining $n-k$ balls in a row, with $n-k+1$ "gaps" (including the first and last gaps);

【Step 3】Insert the $l$ groups of balls from Step 1 into $l$ of the $n-k+1$ gaps, arranging them in a row. The number of ways to insert the balls is the combination of choosing $l$ gaps from $n-k+1$ gaps, i.e., $\mathrm{C}_{n-k+1}^{l}(l \leqslant n-k+1)$;

【Step 4】Number the $n$ balls from left to right as $1,2, \cdots, n$. The numbers corresponding to the $k$ balls inserted in Step 3 are $a_{1}, a_{2}, \cdots, a_{k}$. Thus, the $k$-element subset $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ is a $k$-element subset without isolated elements that satisfies the problem's conditions.
Therefore, the number of $k$-element subsets without isolated elements that satisfy the ordered tuple $\left(k_{1}^{*}, k_{2}^{*}, \cdots, k_{l}^{*}\right)$, i.e., with $l$ segments, each containing $k_{1}^{*}, k_{2}^{*}, \cdots, k_{l}^{*}$ consecutive positive integers, is $\mathrm{C}_{n-k+1}^{l}$.
Using the multiplication principle, the number of $k$-element subsets without isolated elements with $l$ segments is $\mathrm{C}_{k-l-1}^{l-1} \mathrm{C}_{n-k+1}^{l}$.
Since $l \leqslant\left[\frac{k}{2}\right]$, and $l \leqslant n-k+1$, and taking
$$
r=\min \left\{\left[\frac{k}{2}\right], n-k+1\right\},
$$

we have $1 \leqslant l \leqslant r$, i.e., $l$ takes the values $1,2, \cdots, r$.
Using the addition principle, we get
$$
P_{k}=\sum_{l=1}^{r} \mathrm{C}_{k-l-1}^{l-1} \mathrm{C}_{n-k+1}^{l} .
$$