Can you give the general formula( maybe recursive) for the total ways to put n things into m boxes?

Question

n things are similar, i means we can't see any differences between any two of them.
m boxes are similar, too.

Answer 1

This isn't the most easy to understand webpage -- do you know what recurrence relations are? -- but here's an answer:

http://mcraefamily.com/MathHelp/CountingObjectsInBoxes.htm

In your case,

part(m,n) = part(m-1,n-1) + part(m-n,n)

Here's another answer if you know what generating functions are:

http://mathforum.org/dr.math//problems/ali.08.12.01.html

Answer 2

assuming n and m to be small numbers like you have 3 things to be put in 5 boxes.

Assuming 1 obect can be in 1 box.
if you can consider empty boxes as part of your arrangment then it will be 5!=5x4x3x2x1=120 ways of putting 3 things in 5 boxes.
If you do not arrange empty boxes together then it is like
3!=3x2x1=6;

From this a genral formula will be
For empty boxes included it will be m! = m x m-1 x m-2x...x 2 x 1.
if you are not including empty boxes then it is n!.

Now Question is what if boxes are less and things are more then it will always be m!.

Another Question is that if it is possible to put more than 1 thing in a box then this will become even complex but you have to adjust a maximum quanity for each box.

I hope you wont need complex maths here so if you have got the idea then enjoy the statistics of permutation. This is what it is called. Try google search for permutation in statistics and you will get a lot of things to read about.

Answer 3

Well, I know a formula telling you how many ways there are to put n indistinguishable things into m distinguishable boxes.

Total number of ways = C(n+m-1,n), where C denotes combinations and is defined in terms of factorial function as follows: C(n,m) = n!/[m!*(n-m)!]

However, i have no idea whether it is possible to find a formula for the case when the boxes are indistinguishable.

Good luck

Answer 4

m!/ n!(m-n)!

Example:

20 boxes, 3 things
...20.........20!.......20x19x18x...x3x2x1
C... = -------------=----------------------------------
...3.....3! (20-3)!....(3x2x1)(17x16x15x...3x2x1)

So,

...20.... 20x19x18
C... = -------------- = 20x19x3
...3.......3x2x1

Ana

Answer 5

c(n,m)