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(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)

Discussion:

(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 = $\mathbf{ \sum_{i=1}^n i^k }$