#please do not post homework
Kenneth Chan
Math 454
HW 19

1.
egf(AXB)=egf(A)*egf(B)
Where Ais a(k) and B is b(n-k) which gives us A*B=C
the Labels are independent nad distinct
c(n)= the sum of k=0 to n for the binomial(n over k)*akb(n-k)
our goal is to split C(n) is A*B
we can rewrites c(n) as the sum of c(n)/n! *x^n and according to the formula of A*B=C
we can get on the other side of the =the sum of n=0 to infinite summing for k=0 to n substutstitng a(n) into a(k)/k!x^k
and b(N) fir b(n-k)/(n-k)!x^n-k
and we can split them into the sum of k=0 to infinity  for a(k)/k1*x*k which is equal to A and the 
sum of  n-k=0 to infinty for b(n-k)/(n-k)!*x^n-k which gives us B.
Thus C=A*B

2.
ogf=nx^1+(n-1)x^2+(n-2)x^3
efg=a(n)x/n!+a(n-1)x/(n-1)!+a(n-2)x/(n-2)!

3.
(i)10000
(ii)100