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[SOLVED] Simple little combinatorics question.
How many ways are there to distribute 12 identical textbooks to 3 bookshelves?
 one year ago
 one year ago
[SOLVED] Simple little combinatorics question. How many ways are there to distribute 12 identical textbooks to 3 bookshelves?
 one year ago
 one year ago

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KingGeorgeBest ResponseYou've already chosen the best response.1
PS: It's not \[\binom{12}{3}\]
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
i'm quite sure it's 220 tho
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
i was never good at combinatorics =/ 91?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
I believe that is the answer. Good job. Mind explaining how you got that to everyone?
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
i tend to overcomplicate things >.< i think of the bookshelves as:\[x_1,x_2,x_3\]Basically we are asking the question, "How many different integer solutions are there to the equation:\[x_1+x_2+x_3=12\],where each variable must be non negative."
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
That's probably the best way to look at it.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
So to solve that question, I think of the 12 as twelve 1's:\[111111111111\]and im going to pick where to put two +'s. If I place them like this:\[111+11+1111111\]thats like putting 3 books on the first shelf, 2 on the second, and 7 on the third. If I place them like this:\[++111111111111\]thats like placing all the books on the last shelf. Any way you put these book on the shelves, it can be represented by this weird 1's and +'s notation. So now the question is, how man ways can you rearrange twelve 's and two 's? Thats:\[\left(\begin{matrix}14 \\ 2\end{matrix}\right)\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Bingo! This is also called the "stars and bars" problem if anyone wants to look it up.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
typo, twelve 1's and two +'s
 one year ago

robtobeyBest ResponseYou've already chosen the best response.1
http://spot.colorado.edu/~kearnes/F09/distrib_sol.pdf
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
@robtobey did you take that class?
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
i have a question though. My first guess was 3^12, because I was looking at it from each book's perspective, and each book has 3 places it could possibly go. What should tell me, "thats too much"? I mean, i sorta knew that was too much, but why exactly is that the wrong idea?
 one year ago

robtobeyBest ResponseYou've already chosen the best response.1
No. Have not taken a class for decades. Googled the following: "How many ways are there to distribute 12 identical textbooks to 3 bookshelves?"
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
@joemath314159 It's 3^12 if the books are different. You can tell it's too much, because if the textbooks are identical, any permutation of the books will result in the same way to arrange them. @robtobey I actually took that class, so I have that exact sheet in my hands right now. (which is where I got that problem)
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Well, rather the class two years later.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
ah ok. Ive only had a total of one weeks worth of class time in combinatorics. It was a topic in a problem solving course I took a couple of semesters back. We went over this and the principle of inclusionexclusion, which looked really interesting, but my school doesnt offer any full courses on the subject =/
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
I'm taking an actual combinatorics class this fall, and I'm pretty excited for it. This was just part of the twoweek intro in discrete.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
No discrete at my school either ;;
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Maybe it's not called that then. You must have some class that offers an intro to proofs/sets/ other topics that aren't calculus.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
Theres a "foundations of mathematics". That the sets/proofs (although it doesnt do a really good job of that). The Problem Solving course offered is probably the closest we have, but the topics change every onetwo weeks, and can be over anything.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Interesting. I would love to take a class based purely on problem solving.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.4
it is great :) no quizzes, no tests. You turn in 3 proofs a week, of increasing difficulty (if the professor can tell you only took 510 mins to solve the problem, he wont take it). Your grade is based of your proofs and in class participation. The first lecture of the topic is an introduction to the subject, and then the rest of the lectures are spend discussing interesting problems.
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
there's a formula for identical problems: n+r1Cr1
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Nice try @ninhi5 :)
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
nice question, havent taken combination and permutation for ages
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
Elementary application of stars and bars combinatorics. http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
There is a generating function approach to solve this one too.
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
I have already shown that
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
Sorry, I didn't read the whole thread.
 one year ago
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