#ATTENDANCE QUIZ FOR LECTURE 13 of Dr. Z.'s Math454(02) Rutgers University

# Please Edit this .txt page with Answers

#Email  ShaloshBEkhad@gmail.com 
#Subject: p13
#with an attachment called
#p13FirstLast.txt
#(e.g. p13DoronZeilberger.txt)

#Right after finishing watching the lecture but no later than Oct. 20, 2020, 8:00pm


THE NUMBER OF ATTENDANCE QUESTIONS WERE: 6 +1 drawing

PLEASE LIST ALL THE QUESTIONS FOLLOWED, AFTER EACH BY THE ANSWER



#ATTENDANCE Q. #1 for LECTURE 13
#describe the problem that Euler solved regarding 7 bridges

#ANSWER to Q. #1:	

# The Seven Bridges of Koenigsberg problem involved the question if someone could visit all parts of the town by only crossing each of those bridges once (Euler proved that it was not possible, given the arrangement of those bridges,rivers, etc.)




#ATTENDANCE Q. #2 for LECTURE 13
#draw it on a piece of paper with a line segment representing every edge

#ANSWER to Q. #2:

#i drew one on the sketchpad.io website https://gyazo.com/c02943da6b6893edf6bae3fe0f464818




#ATTENDANCE Q. #2 part 2 for LECTURE 13
#if you toss a fair coin 2000 times what is the probability you get exactly 1000 heads (1000 tails)

#ANSWER to Q. #2 part 2:

#every outcome in the sample space has an equal probability, and since we have n=2000 tosses, there are 2^n = 2^2000 outcomes, getting exactly 1000 heads is binomial(2000,1000), so the probability of this outcome is binomial(2000,1000)/2^2000, or roughly 0.018




#ATTENDANCE Q. #3 for LECTURE 13
#if you roll a fair die 6000 times what is the prob that each of the possible outcomes occurs exactly 1000 times

#ANSWER to Q. #3:

#every outcome in the sample space has an equal probability, with n=6000 rolls, there are 6^6000 outcomes, and we do (binomial(6000,1000)*binomial(5000,1000)*binomial(4000,1000)*binomial(3000,1000)*binomial(2000,1000))/6^6000 to get the outcomes where we have 1000 of each, since given 6000 positions for 1000 1's, we then have 5000 for 1000 2's, 4000 for 1000 3's, etc.  Our answer is roughly 7.82 * 10^-10




#ATTENDANCE Q. #4 for LECTURE 13
#if you roll a fair die 6000 times what is the prob that each of the possible outcomes occurs exactly 1000 times

#ANSWER to Q. #4:

#every outcome in the sample space has an equal probability, with n=6000 rolls, there are 6^6000 outcomes, and we do (binomial(6000,1000)*binomial(5000,1000)*binomial(4000,1000)*binomial(3000,1000)*binomial(2000,1000))/6^6000 to get the outcomes where we have 1000 of each, since given 6000 positions for 1000 1's, we then have 5000 for 1000 2's, 4000 for 1000 3's, etc.  Our answer is roughly 7.82 * 10^-10




#ATTENDANCE Q. #5 for LECTURE 13
#pick 5 random facebook friends, for each of them pick 3 friends, for each of the friends of friends pick 3 friends
#label the people picked 1,2,...n, write the graph in our data structure

#ANSWER to Q. #5: initial facebook friends = {1,2,3,4,5}

#(n=1..8)
#G:=[{2,3,4},{1,3,6},{1,2,5},{1,7,8},{3,6,7},{2,5,8},{4,5,8},{4,6,7}]




#ATTENDANCE Q. #6 for LECTURE 13
#look at the cities that border piscataway and for each of them those that border them and again until you get to princeton, construct the graph, find the number of paths from piscataway to princeton, find the actual set using paths by using NuPaths, and by using GFt

#ANSWER to Q. #6: 1=piscataway {2,3,4}  2=plainfield {5,6,7}  5=union {2}  6=elizabeth {2}  7=perth amboy {2,3,9}
#		3=new brunswick {1,2,4,7,8,9}  8=princeton{3,9}  9=sayreville{7,8}, 4=somerville {1,3} (very vague/basic graph)

#G:=[{2,3,4},{5,6,7},{1,2,4,7,8,9},{1,3},{2},{2},{2,3,9},{3,9},{7,8}], will be doing paths of length 2 and/or 3

a) #seq(Paths(G,1,8,i),i=2..3) = {[1, 3, 8]}, {[1, 3, 9, 8], [1, 4, 3, 8]} = {[piscataway, NB,princeton]},{[piscataway,new brunswick, sayreville, princeton], [piscataway,somerville,newbrunswick,princeton]}

b) (Will do nuPaths for many more lenghts, say, 2 to 10)
#seq(NuPaths(G,1,8,i),i=2..3) = 1, 2, 10, 21, 81, 191, 641, 1654, 5132

c)
# seq(coeff(taylor(GFt(G,8,t)[1],t=0,i+1),t,i),i=2..10) = 1, 2, 10, 21, 81, 191, 641, 1654, 5132