(1) Let A = {1, … , k} and B = {1, … , n}. Find the number of maps from A to B .
(2) Define \mathbf{ P_k } be the set of subsets of A. Let f be a map from \mathbf{P_k to B } such that if \mathbf{ U , V \in P_k } then \mathbf{ f(U \cup V) }= \mathbf{\text{max} { f(U) , f(V) } } . Find the number of such functions. (For example if k = 3 and n =4 then answer is 100)


(1) For each member x of set A we have n choices for f(x) in B. Hence the number of functions is mathbf{ n^k }

(2) Claim (i): \mathbf{ f(\phi) } is minimum for any such function f from \mathbf{P_k to B } . This is because \mathbf{ f(A_1) = f(\phi \cup A_1 ) = \text{max} { f(A_1), f(\phi) } } hence \mathbf{f(A_1)} must be larger than \mathbf{ f(\phi) } for any member \mathbf{A_1} of \mathbf{ P_k }

Claim (ii) If we fix the values of the singleton sets then the entire function is fixed. That is if we fix the values of f({1}) , f({2}) , … , f({k}). Since for any member \mathbf{A_1} of \mathbf{P_k , {A_1} } is union of several singleton sets. Hence it’s value is the maximum of the functional values of those singleton sets. For example let \mathbf{ A_1 = (1, 2) } then \mathbf{f(A_1) = f({1}\cup{2}) = \text {max} {f({1}) , f({2}) } }

Claim (iii) f({1}) , … , f({k}) are individually independent of each other.

Now we fix \mathbf{ f(\phi) = i}. Since it is the smallest, the singleton sets map to i to n.
Hence if \mathbf{ f(\phi) = 1} each of the singleton sets have n choices from 1 to n; hence there are \mathbf{ n^k } functions. Similarly \mathbf{ f(\phi) = 2} each of the singleton sets have n-1 choices from 2 to n; hence there are \mathbf{ (n-1)^k } functions.

Thus the total number of functions = \mathbf{ \sum_{i=1}^n i^k }