Number of ways to arrange...?

There are two simple problems here, and two complex problems.

Depending on whether the marbles are distinct and the boxes are distinct.

If the "distinctness" is the same for both it's fairly simple.

If not, then it's rather complex.

Distinct marbles, distinct boxes:

Say the boxes are 5 different colors, and marbles 1,2,3,4 in the red box is different

from those same ones in the blue box:

5 choices of box for each marble, 10 marbles: 5^10 ways to do it.

Non-distinct marbles, non-distinct boxes:

Then it is "number of partitions of 10 into at most 5 parts", which is 30.

See: OEIS sequence A001401: http://www.research.att.com/~njas/sequences/A00140...

The more complex problems are:

Non-distinct marbles, distinct boxes:

For each of the 30 partitions, you have to figure out how many ways it can be done:

10 0 0 0 0 (5! / 4! = 5 ways)

9 1 0 0 0 (5! / 3! = 20 ways)

8 2 0 0 0 (ditto)

7 3 0 0 0 (ditto)

6 4 0 0 0 (ditto)

5 5 0 0 0 (5! / (3! 2!)

and so on including

4 3 2 1 0 (5! ways)

and

3 3 2 1 1 (5! / (2! 2!) ways)

You get a much larger number.

Distinct marbles, non-distinct boxes:

This gets very complex because no matter the approach you take,

there are lots of sub-cases to figure.

You could start with the 30 partitions, but then there are many ways to

construct each one, and then many ways to distribute them to the boxes.

I have to leave this part unsolved.

For reference purposes, here are the 30 partitions:

10

5,5

6,4

7,3

8,2

9,1

4,3,3

4,4,2

5,3,2

5,4,1

6,2,2

6,3,1

7,2,1

8,1,1

3,3,2,2

3,3,3,1

4,2,2,2

4,3,2,1

4,4,1,1

5,2,2,1

5,3,1,1

6,2,1,1

7,1,1,1

2,2,2,2,2

3,2,2,2,1

3,3,2,1,1

4,2,2,1,1

4,3,1,1,1

5,2,1,1,1

6,1,1,1,1

actually only 2 cases have been asked, both of which can be calculated easily

DISTINCT MARBLES, DISTINCT BOXES

------------- ------------- ------------- -----------------

each marble has 5 boxes to choose from, so

# of ways = 5^10

IDENTICAL MARBLES, DISTINCT BOXES

------------- -------------- ------------- ------------------

if m is the # of marbles, and b the # of boxes,

total # of ways = (m+b-1)Cb

for 10 marbles & 5 boxes, it would be

(10+5-1)C10

= 14C10

= 1001

explanation:

----------------

consider the diagram below, m being marbles & | partitions dividing into boxes

m m m | m m | m | m m | m m

indicates 3 marbles in box 1, 2 in box 2, 1 in box 3 , 2 each in boxes 4 & 5

note that only 4 partitions | are needed to divide into 5 parts

the partitions can be placed anywhere, e.g.

| m m | m.... indicates 0 marbles in box 1, 2 in box 2, ....

the total # of ways marbles can be put in boxes

can be obtained by calculating the way partitions

(or hat amounts to the same thing, marbles) can be placed

in ( marbles + partitions) taken together