# How many ways are there to distribute $2$ indistinguishable white and $4$ indistinguishable black balls into $4$ indistinguishable boxes?

How many ways are there to distribute $2$ indistinguishable white and $4$ indistinguishable black balls into $4$ indistinguishable boxes?

If the question was asked as "distinct boxes", I can solve. But now, I am confused. Please show how to solve this type of question explicitly.

• They are just letting you know there is nothing different about the boxes. They are distinct as they are not the same box, but they are identical. – Eleven-Eleven Mar 15 '14 at 19:42

If the answer for distinct box is $N$, then $N/4!$ for indistinquisable box. When the boxs are distinct, let them be A,B,C,D. You may think that you have to put balls into boxes by the order of A,B,C,D one by one, and the permutation is 4! for 4 boxes' orders. When the boxes are the same, 4! is not needed. For this kind of questions, "distinct" ones are generally more directly available than "indistinquisable" ones. The solution always involves "cancle the order".

• Your answer is incorrect. For instance, there are four ways of placing all the balls in one box when the boxes are indistinguishable and one way to place them all in one box when the boxes are indistinguishable. Dividing $4$ by $4!$ yields $1/6$ rather than $1$. – N. F. Taussig Jun 22 '16 at 13:16

Since the bags are indistinguishable, what matters here is the number of balls of each color in the boxes.

There are two distinguishable ways to distribute the two indistinguishable white balls to the four indistinguishable boxes, either both balls are placed in one box or two boxes receive one white ball apiece.

There are five ways to distribute the four indistinguishable black balls to the four indistinguishable boxes. They correspond to the partitions of a set with four elements: \begin{align*} 4 & = 4\\ & = 3 + 1\\ & = 2 + 2\\ & = 2 + 1 + 1\\ & = 1 + 1 + 1 + 1 \end{align*} Note that there are two partitions of a set with two elements: \begin{align*} 2 & = 2\\ & = 1 + 1 \end{align*} We consider cases (a brute force approach).

Case 1: The two white balls are placed in one box. The four black balls are placed in one box.

There are two possible outcomes. Either all six balls are placed in one box or the two white balls are placed in a separate box from the four black balls. \begin{align*} &\{4b, 2w\}\\ &\{4b\}, \{2w\} \end{align*}

Case 2: The two white balls are placed in one box. Three of the black balls are in one box, and the other is placed in a separate box.

There are three possible outcomes. The two white balls are placed in the box with three black balls, the box with one black ball, or a box with no black balls. \begin{align*} &\{3b, 2w\}, \{b\}\\ &\{3b\}, \{b, 2w\}\\ &\{3b\}, \{b\}, \{2w\} \end{align*}

Case 3: The two white balls are placed in one box. Two of the black balls are placed in one box, and the other two black balls are placed in another box.

There are two possible outcomes. The two white balls are placed in a box with two black balls or a box with no black balls. \begin{align*} &\{2b, 2w\}, \{2b\}\\ &\{2b\}, \{2b\}, \{2w\} \end{align*}

Case 4: The two white balls are placed in one box. Two of the black balls are placed in one box, with two other boxes receiving one black ball apiece.

There are three possible outcomes. The two white balls are placed in the box containing two black balls, a box containing one black ball, or the box containing no black balls. \begin{align*} &\{2b, 2w\}, \{b\}, \{b\}\\ &\{2b\}, \{b, 2w\}, \{b\}\\ &\{2b\}, \{b\}, \{b\}, \{2w\} \end{align*}

Case 5: Two white balls are placed in one box. Each of the four black balls is placed in a separate box.

There is one possible outcome. Two white balls are placed in one of the four boxes with a black ball. $$\{b, 2w\}, \{b\}, \{b\}, \{b\}$$

Case 6: The two white balls are placed in separate boxes. All four black balls are placed in one box.

There are two cases. The four black balls are placed in a box with one white ball or no white balls. \begin{align*} &\{4b, w\}, \{w\}\\ &\{4b\}, \{w\}, \{w\} \end{align*}

Case 7: The two white balls are placed in separate boxes. Three black balls are placed in one box, and one black ball is placed in another box.

There are four possible outcomes. Either a white ball is placed in the box with three black balls or it is not. Either a white ball is placed in the box with one black ball or it is not. \begin{align*} &\{3b, w\}, \{b, w\}\\ &\{3b, w\}, \{b\}, \{w\}\\ &\{3b\}, \{b, w\}, \{w\}\\ &\{3b\}, \{b\}, \{w\}, \{w\} \end{align*}

Case 8: The white balls are placed in separate boxes. Two black balls are placed in one box, and the other two are placed in another.

There are three possible outcomes. The white balls are placed in both boxes containing two black balls, exactly one white ball is placed in a box containing two black balls, or no white balls are placed in a box containing two black balls. \begin{align*} &\{2b, w\}, \{2b, w\}\\ &\{2b, w\}, \{2b\}, \{w\}\\ &\{2b\}, \{2b\}, \{w\}, \{w\} \end{align*}

Case 9: The white balls are placed in separate boxes. Two black balls are placed in one box, with two other boxes receiving one black ball apiece.

There are four possible outcomes. If a white ball is placed in the box with two black balls, the other white ball is placed in a box containing one black ball or no black balls. Otherwise, one white ball is placed in each box containing one black ball or one white ball is placed in a box containing one black ball and the other white ball is placed in a box containing no black balls. \begin{align*} &\{2b,w\}, \{b,w\}, \{b\}\\ &\{2b, w\}, \{b\}, \{b\}, \{w\}\\ &\{2b\}, \{b, w\}, \{b, w\}\\ &\{2b\}, \{b, w\}, \{b\}, \{w\} \end{align*}

Case 10: The two white balls are placed in separate boxes. Each of the four black balls is placed in a separate box.

There is one possible outcome. Two boxes receive a black ball ball and a white ball, while the other two receive only a black ball. $$\{b, w\}, \{b, w\}, \{b\}, \{b\}$$

Total: Since the ten cases are mutually exclusive, the answer to your question is found by adding the number of outcomes for each case.
$$2 + 3 + 2 + 3 + 1 + 2 + 4 + 3 + 4 + 1 = 25$$