Question 1210169: In how many ways can 5 balls be placed in 4 boxes, if 2 balls are white and 3 balls are black? (Balls of the same color are indistinguishable. The boxes are indistinguishable.)
Found 3 solutions by CPhill, mccravyedwin, Edwin McCravy:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! This is a tricky problem involving partitions. Since the boxes and the balls of the same color are indistinguishable, we need to consider the different ways to partition the balls into the boxes.
Let's break it down case by case:
**1. Total Balls and Boxes**
* We have 5 balls: 2 white and 3 black.
* We have 4 indistinguishable boxes.
**2. Possible Partitions**
We need to consider the ways to distribute the white balls and black balls separately, and then combine those possibilities.
**Black Balls (3 Balls)**
* **3-0-0-0:** All 3 black balls in one box.
* **2-1-0-0:** 2 black balls in one box, 1 in another.
* **1-1-1-0:** 1 black ball in each of three boxes.
**White Balls (2 Balls)**
* **2-0-0-0:** Both white balls in one box.
* **1-1-0-0:** 1 white ball in each of two boxes.
**3. Combining the Partitions**
Now, we need to consider the combinations of these partitions that result in 5 balls in total.
* **Case 1: (3-0-0-0) Black Balls**
* (3-0-0-0) black and (2-0-0-0) white: (3+2)-0-0-0 = 5-0-0-0
* (3-0-0-0) black and (1-1-0-0) white: (3+1)-1-0-0 = 4-1-0-0
* **Case 2: (2-1-0-0) Black Balls**
* (2-1-0-0) black and (2-0-0-0) white: (2+2)-1-0-0 = 4-1-0-0
* (2-1-0-0) black and (1-1-0-0) white: (2+1)-(1+1)-0-0 = 3-2-0-0
* (2-1-0-0) black and (1-0-0-0) white and (1-0-0-0) white. in this case, the white balls have to go into two boxes, so this case is covered in (2+1)-(1+1)-0-0
* **Case 3: (1-1-1-0) Black Balls**
* (1-1-1-0) black and (2-0-0-0) white: (1+2)-1-1-0 = 3-1-1-0
* (1-1-1-0) black and (1-1-0-0) white: (1+1)-(1+1)-1-0 = 2-2-1-0
**4. Unique Partitions**
Now, let's list the unique partitions we've found:
* 5-0-0-0
* 4-1-0-0
* 3-2-0-0
* 3-1-1-0
* 2-2-1-0
**5. Counting the Partitions**
We have found 5 unique partitions of the 5 balls into 4 indistinguishable boxes.
**Therefore, there are 5 ways to place the 5 balls into the 4 indistinguishable boxes.**
Answer by mccravyedwin(406)  (Show Source):
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
There are 6 non-increasing partitions of 5 with 4 addends 0 through 5. Below
each non-increasing partition are listed the possible distributions of white and
black balls possible in each case.
1. 5 + 0 + 0 + 0 = 5
WWBBB + 0 + 0 + 0 = 5
1 way.
2. 4 + 1 + 0 + 0 = 5
WBBB + W + 0 + 0 = 5
WWBB + B + 0 + 0 = 5
2 ways.
3. 3 + 2 + 0 + 0 = 5
WBB + WB + 0 + 0 = 5
WWB + BB + 0 + 0 = 5
BBB + WW + 0 + 0 = 5
3 ways.
4. 3 + 1 + 1 + 0 = 5
WBB + W + B + 0 = 5
WWB + B + B + 0 = 5
BBB + W + W + 0 = 5
3 ways.
5. 2 + 2 + 1 + 0 = 5
WW + BB + B + 0 = 5
WB + WB + B + 0 = 5
WB + BB + W + 0 = 5
3 ways.
6. 2 + 1 + 1 + 1 = 5
WW + B + B + B = 5
WB + W + B + B = 5
BB + W + W + B = 5
3 ways.
Total = 1 + 2 + 3 + 3 + 3 + 3 = 15 ways
Edwin
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