If `A={x_(1), x_(2), x_(3), x_(4)}` and `B={y_(1), y_(2), y_(3), y_(4)}`, then the number of possible one-one functions defined from A to B

Consider set A = {x_(1), x_(2), x_(3), x_(4), x_(5)} and set B = {y_(1), y_(2), y_(3)} . Function f is defined from A to B. Number of function from A to B such that f(x_(1)) = y_(1) and f(x_(2)) != y_(2) is

If A = {x_(!) , y_(1), z_(1)} and B = {x_(2), y_(2)} , then the number of relations from A to B is

Let A={x_(1),x_(2),x_(3),x_(4)),B={y_(1),y_(2),y_(3),y_(4)} and function is defined from set Ato set B Number ofone- one function such that f(x_(1))!=y_(1) for i=1,2,3,4 is equal to

Consider set A={x_(1),x_(2),x_(3),x_(4)} and set B={y_(1),y_(2),y_(3)} Function f is defined from A to B

The A={x_(1),y_(1),z_(1)} and B={x_(2),y_(2)}, then the number of relations from A to B is

A={x_(1),x_(2),x_(3),x_(4),x_(5)},B={y_(1),y_(2),y_(3),y_(4),y_(5)} A one-one mapping is selected at random from the set of mapping from A to B, the probability that it satisfies the condition f(x_(1))!=y_(1),i=1,2,3,4,5 is

Let A={x_(1),x_(2),x_(3)....,x_(7)},B={y_(1)y_(2)y_(3)} . The total number of functions f:AtoB that are onto and ther are exactly three elements x in A such that f(x)=y_(2) , is equal to