#OK to post homework
#William Wang, 11/12/2020, Assignment 19

#1.
#Suppose that there are two combinatorial familes of objects A and B, and there are m labeled objects in A and n labeled objects in B. Then C x B, is equal to the sum of every ai times every bj times binomial(n+m,2), since we do not want to count the same ai,bj twice. By breaking this up into two sums, we get the sum of m=0 to infinity of a(m)*x`^(m) divided by m!, times the sum of n=0 to infinity of b(n)*x^(n) divided by n!. Thus, the egf of A x B is equivalent to the egf of A times the egf of B, or egf(AxB)=egf(A)*egf(B)

#2.
a:=proc(n): n*(n-1)*(n-1):end:
#i.
#ogf for a: Sum(a(n)*x^(n),n=0..infinity) = Sum(n*(n-1)*(n-2)*x^(n),x=0..infinity) = a(0)+a(1)*x+a(2)*x^(2)+ ...
#ii.
#egf for a: Sum(a(n)*(x^(n))/(n!),n=0..infinity) = Sum((x^(n))/((n-3)!),n=0..infinity)

#3.
#i.
100*100;
 10000
#ii.
coeff(taylor((exp(exp(x) - 1))^2, x = 0, 101), x, 100)*100!;
 86635961604145793709294621186421021577187015751602637085427283871117865337554218116225569701173834817160231371909161518471870