Four identical oranges and six distinct apples (each a different variety) are distributed randomly into five distinct boxes. The probability that each box gets a total of two objects is .

Options

A. `(813)/(109375)`
B. `(162)/(21875)`
C. `(323)/(43750)`
D. `(151)/(21875)`.

Answer

To solve the problem of distributing 4 identical oranges and 6 distinct apples into 5 distinct boxes such that each box gets a total of 2 objects, we can follow these steps:

### Step 1: Calculate the Total Number of Ways to Distribute the Fruits
First, we need to find the total number of ways to distribute the 4 identical oranges and 6 distinct apples into 5 distinct boxes.

1. **Distributing 4 identical oranges**: The number of ways to distribute \( n \) identical items into \( r \) distinct boxes is given by the formula:
\[ \binom{n + r - 1}{r - 1} \]
Here, \( n = 4 \) (oranges) and \( r = 5 \) (boxes):
\[ \text{Ways to distribute oranges} = \binom{4 + 5 - 1}{5 - 1} = \binom{8}{4} = 70 \]

2. **Distributing 6 distinct apples**: Each apple can go into any of the 5 boxes. Therefore, the number of ways to distribute 6 distinct apples is:
\[ 5^6 = 15625 \]

3. **Total ways to distribute fruits**: The total number of ways to distribute both oranges and apples is:
\[ \text{Total ways} = 70 \times 15625 = 1093750 \]

### Step 2: Calculate the Favorable Outcomes
Next, we need to find the number of favorable outcomes where each box gets exactly 2 objects.

#### Case 1: 2 Oranges in 2 Boxes
- Choose 2 boxes to place 2 oranges:
\[ \binom{5}{2} = 10 \]
- The remaining 3 boxes will have 0 oranges.
- The 6 distinct apples need to be distributed in such a way that each of the remaining boxes gets 2 apples. The number of ways to distribute the apples is:
\[ \frac{6!}{2! \times 2! \times 2!} = 90 \]
- Total for this case:
\[ 10 \times 90 = 900 \]

#### Case 2: 2 Oranges in 1 Box, 1 Orange in 2 Boxes
- Choose 1 box for 2 oranges:
\[ \binom{5}{1} = 5 \]
- Choose 2 boxes for 1 orange each:
\[ \binom{4}{2} = 6 \]
- The 6 distinct apples can be distributed in the remaining boxes:
\[ \frac{6!}{2! \times 2! \times 2!} = 90 \]
- Total for this case:
\[ 5 \times 6 \times 90 = 2700 \]

#### Case 3: 1 Orange in 4 Boxes
- Choose 4 boxes to place 1 orange each:
\[ \binom{5}{4} = 5 \]
- The last box will have 0 oranges.
- The 6 distinct apples can be distributed in the remaining boxes:
\[ \frac{6!}{2! \times 2! \times 2!} = 90 \]
- Total for this case:
\[ 5 \times 90 = 450 \]

### Step 3: Total Favorable Outcomes
Now, we sum the favorable outcomes from all cases:
\[ \text{Total favorable outcomes} = 900 + 2700 + 450 = 4050 \]

### Step 4: Calculate the Probability
Finally, we calculate the probability that each box gets a total of 2 objects:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{4050}{1093750} \]

### Step 5: Simplifying the Probability
This fraction can be simplified:
\[ P = \frac{4050}{1093750} \approx 0.0037 \]

### Final Answer
The probability that each box gets a total of two objects is:
\[ \frac{4050}{1093750} \text{ or approximately } 0.0037 \]