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1.
(a)
Using NuW function from ComboProject3.txt
NuW([30, 25, 20], {[0, 0, 1], [0, 1, 0], [1, 0, 0]}) =
same as doing (30+25+20)!/(30!*25!*20!) = 
2478456799816702091914145225486880

(b)
Using NuGW function from ComboProject3.txt
NuGW([30, 25, 20], {[0, 0, 1], [0, 1, 0], [1, 0, 0]}) = 
41512613892711511465063226481480

(c)
Using NuW function from ComboProject3.txt
NuW([30, 25, 20], {[0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1], [2, 2, 2]}) = 
31831650298097356165593942880232160

(d)
edit NuGW function from ComboProj3.txt, where for the conditional if statement, return 0 for cases where x<y or x<z to make sure the walks stay in x>=y Union x>=z like so:

NuGWmod:=proc(Pt,A) local a:
option remember:
if (Pt[1]<0 or Pt[2]<0 or Pt[3]<0 or Pt[1]<Pt[2] or Pt[1]<Pt[3]) then
 RETURN(0):
elif Pt=[0,0,0] then
 RETURN(1):
else
add(NuGWmod(Pt-a,A),a in A):
fi:
end:

NuGWmod([30, 25, 20], {[0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1], [2, 2, 2]}) = 3818966446210163616599115660989544

(e)
ways you can walk in the k-dimensional hyper-cubic lattice can be given by Catalan numbers using this formula:
k=2: (2*n)!/(n!*(n+1)!)
k=3: (2*(3*n)!)/(n!*(n+1)!*(n+2)!)
k=4: (12*(4*n)!)/(n!*(n+1)!*(n+2)!*(n+3)!)
...
(k*n)! * Product(i!/(n+i)!, i=0..k-1)



2.
(a)
words in the alphabet {0,1,2,3} of length 30
This is a case that can be solved with permutations with repetition. There are n=4 different objects in the alphabet where we choose r=30 times since the word is of length 40. This is equivalent to: 
4^30 = 1152921504606846976

(b)
Using functions CompsGF and GFseq from M8.txt
GFseq(CompsGF({2,3,7},x),x,411)[412] = 
18012939919656512113252584133785878056327163962817704791464695718041992634738504938769

(c)
2^(n-1)

(d)
Using function CompsGFcon from M8.txt
CompsGFcon(3,{1},x) = 
(x - 1)*(x^2 + x + 1)/(x^3 + x - 1)

(e)
f:= (x - 1)*(x^2 + x + 1)/(x^3 + x - 1)
evalf(coeff(taylor(f,x,5002), x, 5001)/(coeff(taylor(f,x,5001), x, 5000))) = 
1.465571232



3.
(a)
egf := exp(exp(x)-1);
coeff(taylor(egf,x=0,52),x,51)*51! =
3263983870004111524856951830191582524419255819477

(b)
egf := sum((exp(x)-1)^(3*i+1)/(3*i+1)!, i = 0..infinity);
coeff(taylor(egf,x=0,53),x,51) * 51! = 
1089030250824782148720306369738651363734233026160



4.
(a)
{{1},{2},{3}}
{{1},{2}},{{3}}
{{1},{3}},{{2}}
{{2},{3}},{{1}}
{{1}},{{2}},{{3}}
{{1,2},{3}}
{{1,2}},{{3}}
{{1,3},{2}}
{{1,3}},{{2}}
{{2,3},{1}}
{{2,3}},{{1}}
{{1,2,3}}

(b)
Sum(s[2](n)*x^n/n!, n=0..infinity)
This corresponds to 
egf = exp(exp(exp(x)-1)-1)
1, 3, 12, 60, 358, 2471, 19302, 167894, 1606137, 16733779, 188378402, 2276423485, 29367807524, 402577243425, 5840190914957, 89345001017415, ... 
A000258

(c)
m=11



5.
(a)
Used pn function from M24.txt
pn(1000) = 24061467864032622473692149727991

(b)
Used pnC from M24.txt
pnC(1000,{1},2) = 8635565795744155161506

(c)
Used pnD from M24.txt
pnD(1000,1) = 8635565795744155161506

(d)
Used pnOddDistinctSeq from M24.txt
pnOddDistinctSeq(1000)[1000] = 517035762467311



6.
(a) 
total number of rooted labeled trees = n^n-1 = 31^(31-1) = 
550618520345910837374536871905139185678862401



7.
(a)
BT(4) = 
{ 
  [[[],[]],[[],[]]],
  [[],[[],[[],[]]]], 
  [[],[[[],[]],[]]],
  [[[[],[]],[]],[]],
  [[[],[[],[]]],[]]  
}