#`Please do not post homework`
#`Jiecheng Guo, 09.14.2020, Assignment 2`
#question1
#`if a = 0 and b = 0, then F(a,b,c) = 1. If a = 0 and c = 0, then F(a,b,c) = 1. If b = 0 and c = 0, then F(a,b,c) = 1.`
#`F(a,b,c-1) = F(a+b+c-1)!/(a!*b!*`(c-1)!), F(a,b-1,c) = F(a+b+c-1)!/(a!*(b-1)!*c!), and F(a-1,b,c) = F(a+b+c-1)!/((a-1)!*b!*c!).
#` F(a,b,c)=F(a,b,c-1)+F(a,b-1,c)+F(a-1,b,c)=(a+b+c-1)!/(a!b!(c-1)!)+(a+b+c-1)!/(a!(b-1)!c!)+(a+b+c-1)!/((a-1)!b!c!`
#` (a+b+c-1)!/(a(a-1)!b(b-1)!(c-1)!)+(a+b+c-1)!/(a(a-1)!(b-1)!c(c-1)!)+(a+b+c-1)!/((a-1)!b(b-1)!c(c-1)!)`
#` (a+b+c-1)!/((a-1)!(b-1)!(c-1)!)(1/ab+1/ac+1/bc)`
#` (a+b+c-1)!/((a-1)!(b-1)!(c-1)!)((c+b+a)/abc)`
#` (a+b+c!)/(a!b!c!) `

#question2
#The number of walks in the k-dimensional Manhattan lattice from (0,0,...,0) to (a[i],a[2],...,a[3]) is as the same number of words in the alphabet {1,2,...,k} with the letter i showing up exactly a[i] times because different letters in the alphabet are essentially different dimensions, with additional repetitions of a letter i being like additional steps in the ith direction.

#question3
#` The explicit formula is  (a[1]+...a[k])! / (a[1]!*...*a[k]!)Let S = sum(a[i],i=1..k)`
#` Since a[1] represents the number of occurences of the letter '1'. a[2] `
#` represents the number of occurences of the letter '2', etc.`
#` We can say that the posssible choices for a[1] is binomial (S, a[1]), `
#` and after removing letters '1', a[2] as binomial(S-a[1], a[2]), etc.`
#` Thus we get that for all a[i] from 1 to k, we get that for all occurences of a[i] as `
#`binomial(S,a[1])*binomial`(S-a[1],a[2])*binomial(S-a[1]-a[2],a[3])*...*binomial(S-a[1]-a[2]-a[3]-...-a[k-1], a[k])
#`= S!/((S-a[1])!*`(a[1])!)*(S-a[1])!/((S-a[1]-a[2])!*(a[2])!)*...*(S-a[1]-a[2]-...-a[k-1])!/((S-a[1]-a[2]-...-a[k])!(a[k])!)
#` Since we can cancel out the numerator and denominator for most of this equation, we get`
#` = S!/(a[1]!a[2]!a[3]!...a[k]!)`

#question4
#`from maple code 1 `
 RestrictedWalks :=proc(m,n,s) local W1,W2,w1,w2,curr:

option remember:

if m < 0 or n < 0 then
    RETURN({});
end if:
curr := [m, n]
if member(curr, s) then
    RETURN({});
end if:
if m = 0 and n = 0 then
    RETURN({[]});
end if:
W1 := RestrictedWalks(m - 1, n, s):
W2 := RestrictedWalks(m, n - 1, s):
{seq([op(w1), E], w1 in W1), seq([op(w2), N], w2 in W2)}:
end:

#question5
#from maple code 1
RestrictedNuWalks := proc(m,n,s) local W1,W2,curr:
option remember:
if m < 0 or n < 0 then
    RETURN(0);
end if:
curr := [m, n]:
if m = 0 and n = 0 and member(curr, s) then
    RETURN(0);
end if:
if m = 0 and n = 0 then
    RETURN(1);
end if:
if member(curr, s) then
    RETURN(0);
end if:
W1 := RestrictedNuWalks(m - 1, n, s):
W2 := RestrictedNuWalks(m, n - 1, s):
W1 + W2:
end: