OK to post
1.
a. (30+25+20)!/(30!25!20!)=2478456799816702091914145225486880
b. 
NuGW([30,25,20],{[1,0,0],[0,1,0],[0,0,1]})
41512613892711511465063226481480
c. 31831650298097356165593942880232160
d. 3818966446210163616599115660989544

2.
a.4^30
b.18012939919656512113252584133785878056327163962817704791464695718041992634738504938769
c.
(x-1)/x
d.
(x-1)*(x^2+x+1)/(x^3+x-1)
e.
1.465571232


3.
a.
f:=e^(e^x-1)
SeqBell3(51)[51]
3263983870004111524856951830191582524419255819477
b.
f:=add((exp(x)-1)^(3*i+1)/(3*i+1)!,i=0..51)
coeff(taylor(f,x=0,52),x,51)*51!;
1089030250824782148720306369738651363734233026160
c.
f:=exp(x)-1-add(x^(2*i)/(2*i)!,i=1..25):
f1:=add(f^i/i!,i=0..51):
coeff(taylor(f1,x=0,52),x,51)*51!;
101595492022700597152023270326642584710021120

d.
f:=exp(x)-1-add(x^(2*i)/(2*i)!,i=1..25):
f1:=add(f^(3*i+1)/(3*i+1)!,i=0..51):
coeff(taylor(f1,x=0,52),x,51)*51!;
30143119660583350838086153637731292910753195

4.
a.
k=3 {{{1}},{{2}},{{3}}}
k=2 {{{1},{2}},{{3}}}
{{{1},{3}},{{2}}}
{{{3},{2}},{{1}}}
k=1 {{{1,2,3}}}
{{{3,2},{1}}}
{{{1,2},{3}}}
{{{1,3},{2}}}

b.
egf=exp(exp(x)-1)
coeff(taylor(egf^2/n!,x=0,n+1),x,n)*n!
A001861

c.
coeff(taylor(egf^m/n!,x=0,n+1),x,n)*n!



5.
a.
pn(1000)
24061467864032622473692149727991
b.
The same as distinct partition
8635565795744155161506
c.
pnD(1000,1)
8635565795744155161506
d.
pnDC(1000,1,{1},2)
517035762467311


6.
a.
TreeSeq(31)[31]*31
550618520345910837374536871905139185678862401

b.
SeqRTchild({2},31)[31]/31
78478557180356488765408999200000000

c.
SeqRTchild({1,3,4},31)[31]
388391178300656853181332829972846230000000

d.
SeqRTchildNone({1,3,4},31)[31]
2863465748628395267057869436794752181

7.
a.
[[[],[]],[[],[]]]
[[],[[],[[],[]]]]
[[],[[[],[]],[]]]
[[[[],[]],[]],[]]
[[[],[[],[]]],[]]

b.
There are two cases for complete binary tree since a node can only have two children or none.
Remove the node
Case 1: It is a single vertex so egf is x^0/0!=1
Case 2: It becomes two smaller trees. The order of where two smaller trees are do matter because it can be either left or right child, so egf is B(x)^2
B={} union BxB
Therefore, B(x)=x+B(x)


Modified procedure used:

NuWalks:=proc(m,n,k) local W1,W2,w1,w2,W3,W4,W5:
option remember:

if m<0 or n<0 or k<0 then
 RETURN(0):
fi:


if m=0 and n=0 and k=0 then
 RETURN(1):
fi:

W1:=NuWalks(m-1,n,k):
W2:=NuWalks(m,n-1,k):
W3:=NuWalks(m,n,k-1):
W4:=NuWalks(m-1,n-1,k-1):
W5:=NuWalks(m-2,n-2,k-2):



W1+W2+W3+W4+W5:
end:

RestrictedNuWalks:=proc(m,n,k,S) local W1,W2,W3:
option remember:

if member([m,n,k],S) then
 RETURN(0):
fi:

if m<0 or n<0 or k<0 then
 RETURN(0):
fi:


if m=0 and n=0 and k=0 then
 RETURN(1):
fi:

W1:=RestrictedNuWalks(m-1,n,k,S):
W2:=RestrictedNuWalks(m,n-1,k,S):
W3:=RestrictedNuWalks(m,n,k-1,S):

W1+W2+W3:
end:


NuGW:=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(NuGW(Pt-a,A),a in A):
fi:
end: