In how many ways can three different balls be arranged in nine different boxes in a row such that the number of balls in each box does not exceed 1?

To solve this problem, we need to consider two things: the number of ways to choose the boxes for the balls and the number of ways to arrange the balls in the chosen boxes.

1. Choosing the Boxes: We have 9 boxes and we need to choose 3 of them for the balls. This can be done in "9 choose 3" ways, which is a combination. The formula for combinations is C(n, r) = n! / [(n-r)! * r!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial. So, C(9, 3) = 9! / [(9-3)! * 3!] = 84 ways.

2. Arranging the Balls: We have 3 balls and we need to arrange them in the 3 chosen boxes. This can be done in "3 factorial" ways, which is a permutation. The formula for permutations is P(n) = n!, where n is the number of items. So, P(3) = 3! = 6 ways.

Finally, we multiply the number of ways to choose the boxes by the number of ways to arrange the balls to get the total number of ways. So, the total number of ways is 84 * 6 = 504 ways.