In how many ways can 24 identical white ball and 19 black ball be arranged in rows so that no black balls are together.

To find the number of ways to arrange the white and black balls such that no black balls are together, we can use the concept of permutations.

First, let's consider the arrangement of only the white balls, assuming there are no black balls. Since all the white balls are identical, the number of ways to arrange them is simply 1.

Now, let's consider the black balls. We want to find the number of ways to insert the black balls into the arrangement of white balls such that no two black balls are together.

To visualize this, imagine the white balls as "slots" and we need to insert the black balls into these slots. Since we don't want any two black balls to be together, we need to have at least one white ball between each black ball.

Let's represent the white balls as "_", and the black balls as "B". We can arrange them in the following way, where "B" represents the positions of the black balls:

_ B _ B _ B _ B _ B _ B _ B _

From the above arrangement, we can see that there are 25 available slots for the black balls to be inserted: 24 slots between the white balls, and 1 slot on each end.

Now, we need to choose 19 slots for the black balls out of these 25 available slots. We can do this using combinations. The number of ways to choose 19 slots out of 25 is denoted by C(25, 19), which is calculated as:

C(25, 19) = 25! / (19! * (25-19)!)

Simplifying this, we have:

C(25, 19) = 25! / (19! * 6!)

Using the factorial notation (n!), where n! represents the product of all positive integers from 1 to n, we can calculate the value of C(25, 19).

Therefore, the final answer to the question is the product of the number of ways to arrange the white balls (1) and the number of ways to choose and arrange the black balls (C(25, 19)):

Total number of ways = 1 * C(25, 19)