In how many ways can 12 white and 8 black balls be arranged in a row if no two black balls are together?

Answer

To solve the problem of arranging 12 white balls and 8 black balls in a row such that no two black balls are together, we can follow these steps:

### Step-by-Step Solution:

1. **Arrange the White Balls**:
First, we arrange the 12 white balls in a row. The arrangement of these balls will create gaps where the black balls can be placed.

- Arrangement: W W W W W W W W W W W W

2. **Identify the Gaps**:
After placing the 12 white balls, we can identify the gaps where the black balls can be placed. There will be 13 gaps in total:
- One gap before the first white ball
- One gap between each pair of white balls (11 gaps)
- One gap after the last white ball

So, the gaps can be visualized as follows:
- _ W _ W _ W _ W _ W _ W _ W _ W _ W _ W _

Total gaps = 12 (white balls) + 1 = 13 gaps.

3. **Choose Gaps for Black Balls**:
We need to choose 8 out of these 13 gaps to place the black balls. Since no two black balls can be together, we can only place one black ball in each selected gap.

The number of ways to choose 8 gaps from 13 is given by the combination formula:
\[ \text{Number of ways} = \binom{13}{8} \]

4. **Calculate the Combination**:
Using the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Here, \( n = 13 \) and \( r = 8 \):
\[ \binom{13}{8} = \frac{13!}{8! \cdot (13-8)!} = \frac{13!}{8! \cdot 5!} \]

5. **Simplify the Factorial Expression**:
We can simplify this expression:
\[ \binom{13}{8} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} \]

6. **Calculate the Numerator and Denominator**:
- Numerator: \( 13 \times 12 \times 11 \times 10 \times 9 = 154440 \)
- Denominator: \( 5! = 120 \)

7. **Final Calculation**:
Now, divide the numerator by the denominator:
\[ \frac{154440}{120} = 1287 \]

### Conclusion:
The total number of ways to arrange 12 white balls and 8 black balls in a row such that no two black balls are together is **1287**.

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