<uncletobai> here's where n-3 is coming from
<uncletobai> for every configuration of 4 adjacent spaces
<lama> i'm so slow thats impossible
<uncletobai> you can pick the left-most or first space of the four
<uncletobai> you can have such a 4-space-configuration starting at 1
<lama> uhm
<uncletobai> you can have one starting at 2 etc
<lama> right
<lama> #HERE####
<lama> ##HERE###
<uncletobai> but you cant start close to the end, that is close to n
<lama> ?
<uncletobai> exactly
<uncletobai> you cant have #######HE
<lama> thats on the opposite end of the row
<lama> thats not gonna fit
<uncletobai> exactly
<uncletobai> so you can only go to n-3
<lama> #####HERE
<uncletobai> because the last configuration to fit is n-3, n-2, n-1, n
<lama> 9-3, thats 6
<lama> heh, it works
<uncletobai> what we did is counting startspaces instead of spaces
<uncletobai> ebcause for every 4-space there is exactly one start-space
<uncletobai> so we just count those instead
<lama> so the general principle is to determine the last working configuration?
<uncletobai> and notice that the only start-spaces possible are 1, 2, all the way to n-3
<lama> uhm
<lama> isn't there plenty of start spaces?
<lama> ############################
<uncletobai> if thats about 20 #'s, then yes, 20-3 = 17 start spaces
<uncletobai> the last possible start space is at 17, because the right-most configuration would be 17,18,19,20
<uncletobai> or in your notation ########################HERE
<uncletobai> the H is in position 17
<lama> i understand now that i has to fit, but i wouldn't guess that this is equal to the number of start spaces
<lama> so
<uncletobai> by start spaces i mean the first of the four adjacent spaces
<uncletobai> or the H in your notation
<lama> crap now i'm not sure how to use this
<lama> HERE is four letters so its convinient :D
<lama> so that way i can determine how many adjacent parking spaces are there
<uncletobai> you could number the spaces (s1,s2,...,sn)
<lama> there are n-3 parking spaces
<lama> oh, and total of n choose 4 possibilities, because these are all subsets of {n}
<lama> i don't know how to write it {1,...,n}
<uncletobai> exactly
<lama> :(
<uncletobai> whats wrong? you got it right
<lama> no i didn't
<lama> i would never figured that out myself
<lama> and there are like 50 more exercices like this one
<uncletobai> these things take a lot of practice
<lama> *excersices
<lama> crap i can't even spell that
<lama> yeah, i have two days, thats not enough
<lama> i have a lot of trouble learning something that can't be algorithmized
<uncletobai> preparing for a test?
<lama> yes, finals
<lama> i'm not sure how to solve problems that can't be algorithmized
<lama> meaning that there is no algorithm which could be followed
<uncletobai> https://www.khanacademy.org/math/probability/probability-and-combinatorics-topic
<lama> whoa
<uncletobai> i haven't watched those but they might be helpful
<lama> there are lots of it, i could try probability using combinatorics :D
<lama> its just
<lama> it seems afwul to drop out from colledge after passing some many hard classes
<lama> and then fail on simple combinatorics
<uncletobai> combinatorics is one of the fields where a lot of practice is necessery but once it "clicks" it will be very easy
<lama> but i'm afraid that it won't click in two days
<lama> i have to learn this and also a conditional probability
<lama> and the worst part is, that the exams is actually easy
<lama> *exam
<lama> but i just can't think like that
<uncletobai> try finding a pdf online that has worked solutions
<lama> i can do statistics and such, because i can follow an algorithm
<lama> but there are gonna be only 2 problems on that, and another 3 will be probability
<lama> haha the exercise continues with: what is the probability that at least two parking spaces are next to each other :D
<uncletobai> given that there are four empty spaces?
<lama> and what does the former (4!(n-4)!)/n! says?
<lama> no, its just another question they have
<lama> no
<lama> yes
<lama> :D
<lama> there are still four empty spaces
<uncletobai> ok
<lama> i could try to figure out what is the probability that no 2 spaces are adjacent
<lama> and then negate
<uncletobai> that was my first idea too
<lama> i have some results but i don't know if they are correct
<lama> so i'm trying reading them and understand
<uncletobai> try to get it yourself first
<uncletobai> might be more rewarding
<uncletobai> i have to go now, ill be back in about 20 mins then we can discuss it some more if you want
<lama> ok:)
<lama> thank you for the help
<uncletobai> the problem seems a bit more tricky than the last
<uncletobai> i found a solution online
<uncletobai> http://math.stackexchange.com/questions/488748/simple-probability-question-with-two-colored-balls
<uncletobai> note that this is the same question as yours
<uncletobai> white balls = taken spac es
<uncletobai> black balls = free spaces
<lama> that looks the same
<lama> in my notes there is 1 - n choose 7 / n choose 4, no idea how we go that, it may be wrong
<lama> i'll read the article
<lama> *got that
<lama> *how we got that
<uncletobai> i wonder where that 7 comes from
<lama> exactly
<lama> There are (m+nm)(m+nm) distinguishable arrangements of the balls in a row
<lama> ?
<lama> m+n choose m
<lama> how
<lama> thats just for black balls
<uncletobai> theres a total of m+n balls
<lama> yes
<uncletobai> and m positions where the black balls are
<lama> (m+n)! should give all possible arrangements
<uncletobai> that would be true if the balls are distinguishable
<uncletobai> say if the black balls would be numbered
<uncletobai> then would black_1 black_2 black_3 != black_2 black_1 black_3
<lama> can i divide (m+n)! by the number of identical combinations?
<lama> like (a,b) = (b,a)
<uncletobai> yes you can
<lama> so i would divide by 2
<uncletobai> in these kinds of problems
<uncletobai> it never matters
<lama> but i don't know how many are is that
<lama> *how many is that
<uncletobai> if you label the objects or not
<uncletobai> but you  have to count the same way in the nominator and denominator
<lama> This is a standard stars-and-bars problem
<lama> uuuuuuuuuhm
<lama> stars are
<lama> stares are park places
<lama> wait
<lama> but stars and bars don't care about ordering
<lama> i can have ***|**|******
<lama> but i can't distinguish between the stars
<uncletobai> yeah thats why i suggested not counting orderings
<uncletobai> the first solution in my link
<uncletobai> does it without ordering doesnt he?
<lama> i have to mark stars somehow, so i can keep track on if the stars represent an empty space or not
<uncletobai> so the way i understood it
<uncletobai> call the empty spaces B
<uncletobai> and look at this thing here
<uncletobai> _ B _ B _ B _ B _
<uncletobai> between two Bs you have to put at least one W
<lama> isn't there n-4 '_' spaces?
<lama> yes
<uncletobai> and on the leftmost and rightmost positions you can, but dont have to put at least one W
<uncletobai> in this notation
<uncletobai> we say you can put multiple Ws in the _ position
<lama> leftmost and rightmost must have W
<uncletobai> think about it as inserting there
<lama> what if it starts with B
<uncletobai> no they dont have to
<lama> ok
<uncletobai> this corresponds to not inserting a W to the left or rightmost position
<uncletobai> then it starts with B
<uncletobai> since _ is empty then
<uncletobai> and that position
<lama> _ means empty space? i thought that B is emtpy space
<uncletobai> empty in the sense that there is no white ball there
<lama> _ means 'empty' in a sense that we haven't decided what to put on it
<lama> white ball is non empty space?
<uncletobai> say for example the combination B W B W W B W B W W W W W
<uncletobai> corresponds to _ B _ B _ B _ B _
<uncletobai> where for the leftmost _ we didnt insert any Ws
<uncletobai> but at the rightmost _ we inserted 5 Ws
<lama> ok then
<uncletobai> so the remaining question is
<uncletobai> in how many ways can we insert Ws into _ B _ B _ B _ B _, such that the inner _s have at least one W inserted into them
<uncletobai> and for that question the guy posts a link to the stars and stripes problem on wikipedia :D
<lama> uhm ah
<lama> that doesn't sound too smart does it :D
<lama> well stars and bars are telling me how to distribute stars into boxes, or between bars
<lama> so if i figure out the order of stars and bars it will spit out the number of combinations
<uncletobai> yeah in that case the Bs are the bars
<uncletobai> so you have | | | |
<uncletobai> and BETWEEN the bars there has to be at least one star
<lama> oh
<lama> theorem two at wiki
<lama> bar is empty space
<uncletobai> yeah
<lama> star is occupied space
<lama> so that means
<lama> k = 4
<lama> n = n
<lama> 7+4-1 choose 4
<lama> 10 choose 4?
<uncletobai> im not sure if theorem two is right
<uncletobai> is the right one here
<uncletobai> because we have something more special
<lama> its not?
<uncletobai> we want at least one star between any two bars
<lama> it says that thoerem two allows for no stars between bars
<uncletobai> but we dont care if there is a star at the leftmost position
<uncletobai> yeah but then we would count configurations where two bars adjacent, which we dont want
<lama> oh, i'm sorry we are doing it the otherway around and then negate it, i'm sorry, my head is spinning
<uncletobai> i think i have it
<uncletobai> its way easier than in this post
<uncletobai> no guarantee this is right, but i think it is
<uncletobai> lets say the occupied spaces correspond to *s
<uncletobai> then we have n-4 stars: *****************
<lama> yes
<lama> n-4 stars
<lama> occupied spaces
<uncletobai> and how many _s between and left and right to the stars?
<uncletobai> or better, in how many positions can we insert a bar?
<uncletobai> if there is 6 stars: * * * * * *, there is 7 positions were we can put a bar, right?
<uncletobai> 5 positions between the stars, 1 left and one right to it
<lama> yes
<lama> yes
<uncletobai> so for n-4 stars there is n-3 positions where we can put bars
<lama> number of bars is greater by one than number of stars
<uncletobai> exactly
<lama> this hold from the theorem
<lama> i think
<uncletobai> this holds just by observing :D
<uncletobai> now notice, that if we put bars between the stars
<uncletobai> they will always be seperated
<uncletobai> by a *
<uncletobai> if we dont put 2 in the same place
<uncletobai> so the total number of such configurations
<uncletobai> is just the number of ways we can pick four positions for the bar
<uncletobai> for the bars
<uncletobai> which is binomial(#number of possible positions for the bars, 4)
<uncletobai> which is binomial(n-3, 4)
<uncletobai> so in total the answer should be 1 - binomial(n-3, 4) / binomial(n,4)
<lama> so
<lama> the total of all 4-touples from the set of all empty spaces is n-3 choose 4
<lama> i mean
<lama> thats the total of all configurations where the are no two adjacent spaces
<uncletobai> the number of ways you can put 4 stars between n-4 stars is n-3 choose 4
<uncletobai> exactly
<uncletobai> 4 bars i mean
<lama> its confusing why are we choosing 4
<uncletobai> the bars correspond to empty parking spaces
<uncletobai> we want 4 of them
<lama> ah
<lama> rigjt
<lama> right
<uncletobai> picture n-4 filled parking spaces
<uncletobai> we want to put 4 empty parking spaces between the cars
<uncletobai> there is n-3 positions where we can do that
<uncletobai> picking 4 out of n-3 is     n-3 choose 4
<uncletobai> its as easy as that
<lama> wait, if there are n-4 filled parking spaces then there are only 4 left
<uncletobai> exactly, we have n-4 filled ones and 4 empty ones
<uncletobai> because there is n in total
<lama> i can shuffle those filled anyway i want because they are indistinguishble
<lama> (bad spelling :D)
<lama> there are n-3 ways to arrange 4 empty parking spaces
<uncletobai> adjacent ones
<lama> n-3 adjacent
<lama> n-3 choose 4  = number of configuration of 4 adjacent free spaces
<uncletobai> n-3 choose 4  = number of configuration of 4 non-adjacent free spaces, sorry i got confused up there
<lama> crap
<lama> :D
<lama> i'm really worried
<lama> i'm very stressed that i won't be able to learn all that in time
<lama> if * represents occupied space and | denotes free space and we want distribute | into * in such way that there are no free space adjacent, then all configurations except || are valid
<lama> is that true?
<uncletobai> yeah
<lama> in that case
<uncletobai> but by picking 4 of the spaces wich are seperated by the *s
<uncletobai> this holds automatically
<lama> why not n-4 + 4 - 1 choose n - 4
<uncletobai> how do you get that?
<lama> i accidently closed the browser
<lama> :D
<lama> i mean the page
<uncletobai> want me to paste the chatlog?
<lama> i used the 2nd theorem, but thats wrong, we should use the 1st
<lama> that would be excellent
<lama> please