#OK to post
#Soham Palande, Assignment 19, 11/15/2020



#PART 1

PROOF: egf(A x B) = egf(A) x egf(B)


We suppose that A and B are two combinatorial families and C= A x B.

We can define the attribute to be the "age". 

Then, the number of people with "age" n in A is given by a(n) and similarly the number of people with "age" n in B is given by b(n).

Then, we define C as A x B where the size (or age) of a pair (a,b) is the sum of the ages of a and b in A and B respectively. The labels are disjoint and distinct. 


Then we define c(n) as the number of labeled objects of size (or age) n in C. 
So, first we must choose a size/age for one of the objects. Say we choose k for the a component of the pair. Then there are a(k) labelled objects of age k in A. Since the pair (a,b) in C must have total size/age n, the age of the b component must be n-k. There are b(n-k) labelled objects of size n-k in B. Thus we can pick the objects in a(k)*b(n-k) ways. 

Then, we assign which of the labels go to the first component in nCk ways.

Thus
c(n) = sum(binomial(n,k)*a(k)*b(n-k), k=0..infinity)  ..(1)


Multiplying by x^n/n! and summing from n=0..infinity on both sides of (1), we get

sum(c(n)*x^n/n!, n=0..infinity) = sum(sum(a(k)*(x^k/k!)*b(n-k)*(x^(n-k)/(n-k)!), k=0..n), n=0..infinity)

				Changing the variable of summation to n-k and separating,

				= sum(a(k)/k!, k=0..infinity)*sum(b(n-k)*x^(n-k)/(n-k)!, (n-k)=0..infinity)

Thus, we get that egf(C)=egf(A)*egf(B)
		  egf(A x B)= egf(A) * egf(B)




#PART 2


Find egf of a(n)=n*(n-1)*(n-2) (n ≥ 0)

f:=sum(n*(n-1)*(n-2)*(x^n)/n!,n=0..infinity)
f := x^3*exp(x)


#PART 3


(i) 

Let X(n) be {[SP1,SP2]} where the size of SP1 + size of SP2 = n 
The labels are {1,...,100}
We want to find the size of X(100)

coeff(taylor(exp(exp(x)-1)*exp(exp(x)-1),x=0,101),x,100)*100!

86635961604145793709294621186421021577187015751602637085427283871117865337554218116225569701173834817160231371909161518471870

(ii)

The above number divided by 2 since in a set the order does not matter. 

So, 

coeff(taylor(exp(exp(x)-1)*exp(exp(x)-1),x=0,101),x,100)*100! / 2