I have a hard time solving this question, for some reason I can't get my head around it. Here it goes:
There are 3 different zones and 6 different type of object of which there are 2 object of each type (12 objects in total). How many different ways can these objects be distributed among the zones?
- The ordering of the objects within the zones are not significant but the ordering of the zones are
- each object must be within a zone
- there is no limitation on how many objects a zone can contain and empty zones are permitted
my brute-force approach:
step 1:
find all the combinations of how many objects can a zone have:
z1 z2 z3
12 0 0
11 1 0
10 2 0
10 1 1
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 1
5 4 2
5 3 3
4 4 4
step2: find how many different ways the object types can be distrubuted among these. I have actually solved it for 2 types of objects but not 6 because it gets way too complex for the 6 object types.
Best Answer
Let's begin with a simpler question:
Let $x_i$ be the number of objects placed in the $i$th zone, where $1 \leq i \leq 3$. Then $$x_1 + x_2 + x_3 = 2$$ is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of $3 - 1 = 2$ addition signs in a row of two ones. For instance, $$1 1 + +$$ corresponds to the solution $(2, 0, 0)$, while $$+ 1 + 1$$ corresponds to the solution $(0, 1, 1)$. The number of such solutions is $$\binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = 6$$ since we must choose which $3 - 1 = 2$ of the $2 + 3 - 1 = 4$ positions required for two ones and two addition signs will be filled with addition signs.
As Jaap Scherphuis pointed out in the comments, each type of object can be distributed independently of the others. Since the two objects of each of the six types of objects can be distributed to the three zones in six ways, there are $6^6$ possible distributions.