How many functions satisfy the property $f(i)Let $F$ be the set of one-to-one functions from the set $\{1, 2, ….., n\}$ to the set $\{1, 2, ….., m\}$ where $m\geq n\geq 1$.


*

*How many functions are members of $F?$

*How many functions $f$ in $F$ satisfy the property $f(i)=1$ for some $i,
        1 \leq i \leq n?$

*How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for some
        $1 \leq i \leq j \leq n$?

Somewhere it explained as :


*

*A function from $A$ to $B$ must map every element in $A$. Being one-one, each element must map to a unique element in $B$. So, for $n$ elements in $A$, we have $m$ choices in $B$ and so we can have $^m\mathbb{P}_n$ functions.

*Continuing from $(1)$ part. Here, we are forced to fix $f(i) = 1$. So, one element from $A$ and $B$ gone with $n$ possibilities for the element in $A$ and $1$ possibility for that in $B$, and we get $n \times ^{m-1}\mathbb{P}_{n-1}$ such functions.

*$f(i) < f(j)$ is satisfied by all such functions as we are considering only one-one functions. So, this answer will be same as in $(1)$.



I've understood well part $(1)$ and part $(2)$ . I'm not getting explanation of part $(3)$.
How is it satisfied by one to one function , and how many such functions are there ?

The "somewhere" explanation is not correct. For a counterexample, take $n=m=2$. Then the function
$$ f = \{ 1\mapsto 2, 2\mapsto 1 \} $$
is in $F$ but does not satisfy the property in (3).
There are in general $\binom mn$ strictly decreasing functions that are in $F$ but don't satisfythe condition.
