> #Weiji Zheng, 09/27/2020, Assignment 5
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> #OK to post homework
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> 
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> #Q1
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> #Prove p(n) = n*p(n-1) where P(n) is broken-up to sets
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> #P(1)=[{1}]
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> #P(2)=[{12},{21}]
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> #P(3)=[{123},{132},{213},{231},{312},{321}]
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> #We have P_j[n]={All permutations pi in P(n) such that that pi[1]=j, j=1..n}
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> #For example, P_1[2] = All permutations pi in P(2) such that that pi[1]=1
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> 
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> #So we have
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> #P_1[3] = [{312},{321}] IN BIJECTION WITH [{12},{21}] P[2]
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> #P_2[3] = [{132},{231}] IN BIJECTION WITH [{12},{21}] P[2]
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> #P_3[3] = [{123},{213}] IN BIJECTION WITH [{12},{21}] P[2]
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> #P_j[n] IS IN BIJECTION WITH P[n-1]
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> 
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> #Then
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> #|P_j[n]| = |P_[n-1]| for j = 1,2,3,...,n
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> #|P[n]| = |P_1[n] union P_2[n] union ... P_n[n]| = |P_1[n]| + ... +|P_n[n]|
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> #|P[n]| = |P[n-1]|+ ... + |P[n-1]| = n*|P(n-1)| = n*(n-1)*|P[n-2]|=...
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> #Since p[0] = 1,we get
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> #p[n] = n*p[n-1] = n!
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> 
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> #Q2
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> #P(1;1,2,3) = {1} p(1)=1
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> #P(2;1,2,3) = {11,2} p(2) = 2
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> #P(3;1,2,3) = [3,{1+P(2)},{2+P(1)}] = [3,111,12,21} p(3) = 4
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> #P(4;1,2,3) = [{1+P(3)},{2+P(2)},{3+P(1)}] = [12,1111,112,121,22,211,3] p(4) = 7
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> #So, for P(n;1,2,3),we have p(n) = |P(n;1,2,3)| = 
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> #p(n-1)+p(n-2)+p(n-3)
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> 
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> #Then for p(n;a1,a2,a3) we can conclude p(n) = p(n-a1)+p(n-a2)+p(n-a3) where p(n) = 1 when n = 0 and p(n) = 0 when n < 0
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> 
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> #Q3
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> #Write a Maple procedure (don't forget to use option remember)SEQP(a1,a2,a3,N) That inputs a1,a2,a3, as above, and a positive integer N, and outputs the first N+1 terms, starting at n=0 of p(n,a1,a2,a3), n=1..N
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> 
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> #Sorry, Dr.Z, it's so hard.
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> 
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> #Q4
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> 
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> #this person is in C and L, not in S an K
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> #So
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> #Line1
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> #this person is included for 2 times -> |C| and |L|
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> #Line2
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> #this person is excluded for 1 times -> |C interset L|
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> #Line3
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> #this person is included for 0 time
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> #Line4
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> #this person is excluded for 0 time
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> 
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> 
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> #this person is in C and L, S an K
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> #So
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> #Line1
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> #this person is included for 4 times -> |C| and |L| and |S| and |K|
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> #Line2
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> #this person is excluded for 6 times ->  |C interset S|,|C interset L|,|C interset K|,|S interset L|,|S interset K|,|L interset K|
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> #Line3
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> #this person is included for 4 time ->  |C interset S intersect L|,|C interset S intersect K|,|C interset K intersect L|,|S interset K intersect L|
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> #Line4
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> #this person is excluded for 1 time -> |C intersect S intersect L intersect K|
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> 
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