Question 2 Given positive integers $m, n$ satisfying $m \geqslant n \geqslant 3$. Find the number of injections $f:\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, m\}$, such that there is a unique $i \in\{1,2, \cdots, n-2\}$, satisfying
$$
f(i)>f(i+1)>f(i+2) \text {. }
$$

For other $j \in\{1,2, \cdots, n-1\} \backslash\{i, i+1\}$, it holds that
$$
f(j)<f(j+1)
$$
Let the image set of the mapping $f$ be
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} .
$$

Then there are $\mathrm{C}_{m}^{n}$ such image sets.
Without loss of generality, let $A=\{1,2, \cdots, n\}$.
We now calculate the number of one-to-one mappings $f$ :
$$
\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, n\}
$$

such that there is a unique $i \in\{1,2, \cdots, n-2\}$ satisfying equation (1), and for all other
$$
j \in\{1,2, \cdots, n-1\} \backslash\{i, i+1\},
$$

equation (2) holds.
$$
\begin{array}{l}
\text { Then } f(1)f(i+1)>f(i+2), \\
f(i+2)<f(i+3)<\cdots<f(n) .
\end{array}
$$

Since there are $i+1$ numbers smaller than $f(i)$, $f(i)$ can only take the values $i+2, i+3, \cdots, n$.
$$
\text { When } f(i)=i+2 \text {, in the set }\{1,2, \cdots, i+1\}
$$

we choose $i-1$ numbers and arrange them in increasing order as
$$
f(1), f(2), \cdots, f(i-1) \text {, }
$$

the smaller of the two remaining numbers in this set is assigned to $f(i+2)$, and the other to $f(i+1)$. The remaining numbers in the set $A=\{1,2, \cdots, n\}$ are arranged in increasing order as
$$
f(i+3), f(i+4), \cdots, f(n) \text {. }
$$

Thus, there are $\mathrm{C}_{i+1}^{i-1}$ such one-to-one mappings.
When $f(i)=i+3$, in the set $\{1,2, \cdots, i+2\}$ we choose $i-1$ numbers and arrange them in increasing order as
$$
f(1), f(2), \cdots, f(i-1) \text {, }
$$

the smallest of the three remaining numbers in this set is assigned to $f(i+2)$, and one of the other two is assigned to $f(i+1)$. The remaining numbers in the set $A=\{1,2, \cdots, n\}$ are arranged in increasing order as
$$
f(i+3), f(i+4), \cdots, f(n) \text {. }
$$

Thus, there are $\mathrm{C}_{i+1}^{i-1} \mathrm{C}_{2}^{1}$ such one-to-one mappings.
And so on.
Finally, when $f(i)=n$, in the set $\{1,2, \cdots, n-1\}$ we choose $i-1$ numbers and arrange them in increasing order as
$$
f(1), f(2), \cdots, f(i-1),
$$

the smallest of the $n-i$ remaining numbers in this set is assigned to $f(i+2)$, and one of the other $n-i-1$ is assigned to $f(i+1)$. The remaining numbers in the set $A=\{1,2, \cdots, n\}$ are arranged in increasing order as
$$
f(i+3), f(i+4), \cdots, f(n) \text {. }
$$

Thus, there are $\mathrm{C}_{n-1}^{i-1} \mathrm{C}_{n-i-1}^{1}$ such one-to-one mappings.
Therefore, when $i$ is given, the number of one-to-one mappings $f:\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, n\}$ that satisfy the conditions is
$$
\begin{array}{l}
\sum_{k=1}^{n-i-1} k \mathrm{C}_{i+k}^{i-1} \\
=\sum_{k=1}^{n-i-1}(i+k+1) \mathrm{C}_{i+k}^{i-1}-\sum_{k=1}^{n-i-1}(i+1) \mathrm{C}_{i+k}^{i-1} \\
=i \sum_{k=1}^{n-i-1} \mathrm{C}_{i+k+1}^{i}-(i+1) \sum_{k=1}^{n-i-1} \mathrm{C}_{i+k}^{i-1} \\
=i\left(\mathrm{C}_{n+1}^{i+1}-i-2\right)-(i+1)\left(\mathrm{C}_{n}^{i}-i-1\right) \\
=(i+1)\left(\mathrm{C}_{n+1}^{i+1}-\mathrm{C}_{n}^{i}\right)-\mathrm{C}_{n+1}^{i+1}+1 \\
=(i+1) \mathrm{C}_{n}^{i+1}-\mathrm{C}_{n+1}^{i+1}+1 .
\end{array}
$$

Thus, the number of one-to-one mappings $f$ :
$$
\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, n\}
$$

that satisfy the conditions is
$$
\begin{array}{l}
\sum_{i=1}^{n-2}\left[(i+1) \mathrm{C}_{n}^{i+1}-\mathrm{C}_{n+1}^{i+1}+1\right] \\
=\sum_{i=1}^{n-2}(i+1) \mathrm{C}_{n}^{i+1}-\sum_{i=1}^{n-2} \mathrm{C}_{n+1}^{i+1}+n-2 \\
=\sum_{i=1}^{n-2} n \mathrm{C}_{n-1}^{i}-\sum_{i=1}^{n-2} \mathrm{C}_{n+1}^{i+1}+n-2 \\
=n\left(\sum_{i=0}^{n-1} \mathrm{C}_{n-1}^{i}-2\right)-\left(\sum_{i=0}^{n+1} \mathrm{C}_{n+1}^{i}-2 n-4\right)+n-2 \\
=n\left(2^{n-1}-2\right)-\left(2^{n+1}-2 n-4\right)+n-2 \\
=(n-4) 2^{n-1}+n+2 .
\end{array}
$$

Therefore, the number of one-to-one mappings $f$ :
$$
\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, m\}
$$

that satisfy the conditions is
$$
\mathrm{C}_{m}^{n}\left[(n-4) 2^{n-1}+n+2\right] .
$$