In how many ways can one arrange seven white and five black balls in a line in such a way that there are no two neighboring black balls?
1. **Initial Placement of Black Balls:**
   - We need to place the five black balls such that no two black balls are adjacent. To achieve this, we can place the black balls with at least one white ball between each pair.
   - Consider the arrangement of black balls as follows: B _ B _ B _ B _ B. Here, the underscores represent the positions where white balls can be placed to ensure no two black balls are adjacent.

2. **Counting the Positions for White Balls:**
   - There are 6 positions (underscores) where we can place the white balls: one before the first black ball, four between the black balls, and one after the last black ball.
   - We have 7 white balls in total, and 4 of them are already used to separate the black balls. Therefore, we have 3 white balls left to place in these 6 positions.

3. **Distributing the Remaining White Balls:**
   - We need to distribute the remaining 3 white balls into the 6 available positions. This is a classic "stars and bars" problem where we need to find the number of ways to distribute \(n\) indistinguishable items (white balls) into \(k\) distinguishable bins (positions).

4. **Using the Stars and Bars Theorem:**
   - The number of ways to distribute \(n\) indistinguishable items into \(k\) distinguishable bins is given by the binomial coefficient \(\binom{n+k-1}{k-1}\).
   - Here, \(n = 3\) (remaining white balls) and \(k = 6\) (positions), so we need to calculate \(\binom{3+6-1}{6-1} = \binom{8}{5}\).

5. **Calculating the Binomial Coefficient:**
   \[
   \binom{8}{5} = \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56
   \]

Thus, the number of ways to arrange seven white and five black balls in a line such that no two black balls are adjacent is \(56\).

The final answer is \(\boxed{56}\).