How many functions satisfy the property $f(i)<f(j)$ for some $1 \leq i \leq j \leq n$? 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 $f(i)&lt;f(j)$ 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.
$f(i) &lt; f(j)$ 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.
answer of c) is actually C(m,n) because you have to just select any n number from m which can be done in C(m,n) ways, and coming to the arrangement, that chosen n numbers should be in strictly increasing , so you have just 1 way to arrange them. Hence if you do selection followed by arrangement it will be C(m,n) * 1, which will be simply C(m,n)