How many one-to-one correspondences exist between two sets with

Question

A) 8 elements each?

B) n-1 elements each?

Answer 1

A) 8*7*6*5*4*3*2*1

Let's say that the first set is {a,b,c,d,e,f,g,h} and the second set is {1,2,3,4,5,6,7,8}. a can be grouped in 8 different ways, afterwards, if you've already grouped a then that leaves b with 7 elements to choose from and then c with 6 elements to choose from and so on.

B) (n-1)! = (n-1)*(n-2)*(n-3)... *3*2*1

I believe this is the answer. The operation works in a similar way to the previous problem. If one element is grouped with another element from the second set then the next element would have one less option with which to group with, hence, n-1, n-2, n-3...