How many ways are there to put 4 b***s in 3 boxes if the b***s are not distinguishable and neither are the boxes?
+1038 If the boxes were distinguishable then your solution would be correct. In effect, this is distributing indistinguishable b***s (k) into distinguishable boxes (N), forming a combination of size k, taken from a set of size n.
$$\displaystyle \left( {\begin{array}{*{20}c} 4 \\3 \\ \end{array}} \right)\; = \; \dfrac{(4)!}{3!(4-3)!} \; = \;\hspace{5pt} \Text {4 \ ways} \\\$$
When both b***s and boxes are indistinguishable, empty boxes are forbidden, if they were allowed then they are distinguishable, and the above formula would apply.
The reason for this is the mathsgod said so. Mathsgod, please come and verify the veracity of this statement.
These types of questions are not intuitive. Consider your statement: “All the b***s and all the boxes are identical . . .” Though it would seem so, this is not the same as indistinguishable in a mathematical sense.
If identical boxes are present at the same time then they are identifiable via their positions in space. Like many hypothetical math questions, it becomes difficult -- sometimes impossible-- to create a real situation where these concepts can be examined. Many math realities exist only in the mathematics and cannot exist in our normal space-time. Torricelli's trumpet is a classic example of this.
This question can exist in normal space but probably not with boxes. An approximation might be to place the boxes behind a frosted glass that distorts their size, shape and position in space. This approximation still falls short. To be true to the question, the boxes cannot exist until the b***s are placed in them. This is why it is a partition question. There are no empty partitions because the b***s themselves make the partitions.
As complex as this may seem, it becomes more so when the b***s are distinguishable. These are royal ball-busters. Sterling numbers of the second kind are used for these partitions.
+23254 I think that there are 15 ways ... call the boxes A, B, and C.
You could put all four b***s into either box A, or box B, or box C ---> 3 ways.
You could put three ball into one box and the other ball into another box:
-- 3 into A, 1 into B, 0 into C; or 3 into A, 0 into B, 1 into C; or 1 into A, 3 into B, 0 into C; or 1 into A, 0 into B, 3 into C; or 0 into A, 3 into B, 1 into C; or 0 into A, 1 into B, 3 into C ---> 6 ways
You could put two b***s into one box and the other two b***s into another box:
-- 2 into A, 2 into B, 0 into C; or 2 into A, 0 into B, 2 into C; or 0 into A, 2 into B, 2 into C ---> 3 ways.
You could put two b***s into one box and one ball into each of the other two boxes:
-- There are three boxes that could get the two b***s ---> 3 ways
Total: 3 + 6 + 3 + 3 = 15 ways.
+1038
The partition function is defined as the number of k-element partitions of N.
In this case, the b***s are the k-element and the boxes are N.
Partitions of 4 constraint of 3. Sometimes denoted as Part(4,3)
$$\displaystyle \text {List: Part(4,k)}
4=4
2+2=4
2+1+1 = 4 \leftarrow \text {This is the only solution}
1+3=4$$
There is only one (1) way to Partition 4 b***s into 3 boxes
Wolframalpha scripted link :
http://www.wolframalpha.com/input/?i=partitions+of+4+into+3+parts
Wolfram script: “partitions of 4 into 3 parts"
Wolframalpha will return the count and list the partition distributions.
+118723 I just counted the posibilities :/
All the ball and all the boxes are identical so all that matters is how many b***s are in each box
4,0,0
3,1,0
2,2,0
2,1,1
and that is it.
4 ways. :)
Nauseated's solution looks a bit too techo for me. :/
+1038 If the boxes were distinguishable then your solution would be correct. In effect, this is distributing indistinguishable b***s (k) into distinguishable boxes (N), forming a combination of size k, taken from a set of size n.
$$\displaystyle \left( {\begin{array}{*{20}c} 4 \\3 \\ \end{array}} \right)\; = \; \dfrac{(4)!}{3!(4-3)!} \; = \;\hspace{5pt} \Text {4 \ ways} \\\$$
When both b***s and boxes are indistinguishable, empty boxes are forbidden, if they were allowed then they are distinguishable, and the above formula would apply.
The reason for this is the mathsgod said so. Mathsgod, please come and verify the veracity of this statement.
These types of questions are not intuitive. Consider your statement: “All the b***s and all the boxes are identical . . .” Though it would seem so, this is not the same as indistinguishable in a mathematical sense.
If identical boxes are present at the same time then they are identifiable via their positions in space. Like many hypothetical math questions, it becomes difficult -- sometimes impossible-- to create a real situation where these concepts can be examined. Many math realities exist only in the mathematics and cannot exist in our normal space-time. Torricelli's trumpet is a classic example of this.
This question can exist in normal space but probably not with boxes. An approximation might be to place the boxes behind a frosted glass that distorts their size, shape and position in space. This approximation still falls short. To be true to the question, the boxes cannot exist until the b***s are placed in them. This is why it is a partition question. There are no empty partitions because the b***s themselves make the partitions.
As complex as this may seem, it becomes more so when the b***s are distinguishable. These are royal ball-busters. Sterling numbers of the second kind are used for these partitions.
