Question

[SOLVED] Simple little combinatorics question.

How many ways are there to distribute 12 identical textbooks to 3 bookshelves?

Answer 1

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

Answer 2

220

Answer 3

Nope.

Answer 4

1320?

Answer 5

i'm quite sure it's 220 tho

Answer 6

i was never good at combinatorics =/ 91?

Answer 7

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

Answer 8

what's the answer?

Answer 9

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."

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

typo, twelve 1's and two +'s

Answer 14

http://spot.colorado.edu/~kearnes/F09/distrib_sol.pdf

Answer 15

@robtobey did you take that class?

Answer 16

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?

Answer 17

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

Answer 18

@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)

Answer 19

Well, rather the class two years later.

Answer 20

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 =/

Answer 21

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.

Answer 22

No discrete at my school either ;;

Answer 23

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.

Answer 24

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.

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

guess i'm too late

Answer 29

Nice try @ninhi5 :)

Answer 30

nice question, havent taken combination and permutation for ages

Answer 31

Elementary application of stars and bars combinatorics.

http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

Answer 32

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

Answer 33

I have already shown that

Answer 34

Sorry, I didn't read the whole thread.