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OpenStudy (kinggeorge):

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

OpenStudy (kinggeorge):

PS: It's not \[\binom{12}{3}\]

OpenStudy (anonymous):

220

OpenStudy (kinggeorge):

Nope.

OpenStudy (anonymous):

1320?

OpenStudy (anonymous):

i'm quite sure it's 220 tho

OpenStudy (anonymous):

i was never good at combinatorics =/ 91?

OpenStudy (kinggeorge):

I believe that is the answer. Good job. Mind explaining how you got that to everyone?

OpenStudy (anonymous):

what's the answer?

OpenStudy (anonymous):

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."

OpenStudy (kinggeorge):

That's probably the best way to look at it.

OpenStudy (anonymous):

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:\[\left(\begin{matrix}14 \\ 2\end{matrix}\right)\]

OpenStudy (kinggeorge):

Bingo! This is also called the "stars and bars" problem if anyone wants to look it up.

OpenStudy (anonymous):

typo, twelve 1's and two +'s

OpenStudy (kinggeorge):

@robtobey did you take that class?

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

No. Have not taken a class for decades. Googled the following: "How many ways are there to distribute 12 identical textbooks to 3 bookshelves?"

OpenStudy (kinggeorge):

@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)

OpenStudy (kinggeorge):

Well, rather the class two years later.

OpenStudy (anonymous):

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 inclusion-exclusion, which looked really interesting, but my school doesnt offer any full courses on the subject =/

OpenStudy (kinggeorge):

I'm taking an actual combinatorics class this fall, and I'm pretty excited for it. This was just part of the two-week intro in discrete.

OpenStudy (anonymous):

No discrete at my school either ;;

OpenStudy (kinggeorge):

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.

OpenStudy (anonymous):

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 one-two weeks, and can be over anything.

OpenStudy (kinggeorge):

Interesting. I would love to take a class based purely on problem solving.

OpenStudy (anonymous):

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 5-10 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.

OpenStudy (anonymous):

there's a formula for identical problems: n+r-1Cr-1

OpenStudy (anonymous):

guess i'm too late

OpenStudy (kinggeorge):

Nice try @ninhi5 :)

OpenStudy (anonymous):

nice question, havent taken combination and permutation for ages

OpenStudy (anonymous):

Elementary application of stars and bars combinatorics. http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

OpenStudy (anonymous):

There is a generating function approach to solve this one too.

OpenStudy (anonymous):

I have already shown that

OpenStudy (anonymous):

Sorry, I didn't read the whole thread.

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