n(X)=p and n(Y)=q then the no. of function from X-> Y is q^p , how do u prove this ?
Show your work so far and we can be of more help. As a hint, consider an element of X; how many choices are there for its image?
Ok , Here is what I've tried to do ,
the the function with max. no. of elements from these 2 sets should be a many-one on-to function right ? The no. of elements in that set(function) should be p*q(if the no of elements in 1st set is p and 2nd is q) , and i think that all other functions from these 2 sets should be subsets of this.
Considering this as a relation (and not a function) the no. subsets should be 2^mn . but for this relation to be a function there should be all the elements of set X in the ordered pairs which are the elements of the relation .
But using this logic and using cobinations I cant seem to get to the result q^p
You might find it easier to start with my previous suggestion: Take an element of X; how many choices are there for its image in Y?
Oh , didn't realize that this was this simple :( , so the no. chances of images for an element is q ,so the total no. of chances is q*q*q.....*q p times . Thanks for the help
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