#OK to post homework
#Tianyi Liu, Nov15, Assignment 19

1.
Suppose there are two cominatorial familes A and B of size n and we define C as AxB where the labels are 
disjoint and distinct. We define the size of a pair as the sum of components in A and B. We get c(n)=sum 
of (n choose k)*a(k)*b(n-k). The size of c would be k of a plus n-k of b which equals n. There are n choose
k ways of choosing which goes to a component. Then we pick objects for a and b from A and B. Multiplying 
both sides by the concept of egf x^n/n! we get sum from n=0 to infinity c(n)*x^n/n!=sum from n=0 to infinity 
sum of k=0 to n a(k)*x^k/k!*b(n-k)*x^(n-k)/(n-k)!
=(sum of k=0 to n a(k)*x^k/k!)*(sum of n-k=0 to n b(n-k)*x^(n-k)/(n-k)!)
=egf(C)=egf(AxB)=egf(A)*egf(B)

2.
OGF:6*x^3/(x-1)^4
EGF:x^3*exp(x)

3.
i. 86635961604145793709294621186421021577187015751602637085427283871117865337554218116225569701173834817160231371909161518471870
ii.43317980802072896854647310593210510788593507875801318542713641935558932668777109058112784850586917408580115685954580759235935