I've considered an example , with n=3(1,2,3) and m=4(1,2,3,4)

for strictly increasing function , if I've mapped (1,2) then for element 2 from set A , I can't map (2,2) since it is one to one , and also I can't map (2,1) because it can't satisfy the property f(i)<f(j) , i.e. 2 !<1 , so element 2 should be map in remaining set element except 1 and 2 so, I've mapped with element(2,3) { I can map with other element of set , but ,we should remember the satifyng property and property of a function.}

Similary , for element 3 , I can't map with below with element 3 of set B , so , remaining number elements is 4 only . so , it should be (3,4).

final mapping ,

example :

A(1,2,3) B(1,2,3,4)

for the satsfying condition f(i)<f(j) .where i , j are from set A and f(i) , f(j) from set B

Total number of such functions are :

1.{(1,1) ,(2,2), (3,3)}

2.{(1,1),(2,2),(3,4)}

3.{(1,1),(2,3),(3,4)}

4.{(1,2),(2,3),(3,4)}

(1,2,3),(1,2,4),(1,3,4),(2,3,4) , is similar to choose (we can see here , odere is not matter) 3 element from 4 element

Only 4 such possible functions .

So , the possible functions are choose n element from m element , i.e., mCn