“There are 5 red balls and 4 red balls. In how many ways can these balls be arranged in a row such that no red balls are adjacent?”

Answer:There are 24 ways to arrange the 5 red balls and 4 blue balls in a row such that no red balls are adjacent.Step-by-step explanation:To solve this problem, we need to arrange the red and blue balls in such a way that no red balls are adjacent. Let’s assume there are 5 red balls and 4 blue balls. We can use the concept of inserting gaps between the balls.Total Balls: There are 9 balls in total (5 red and 4 blue).Insert Gaps for Non-Adjacent Red Balls: To ensure no red balls are adjacent, we need to place 4 blue balls in such a way that they create 5 gaps (since there are 5 red balls). These gaps will be the positions where the red balls can be placed.Arrange Blue Balls: The blue balls can be arranged in any order. Let’s consider one possible arrangement for simplicity:Blue balls: B B B BCreate Gaps: Place the blue balls in a sequence to create gaps:B _ _ _ B _ _ _ B _ _ BPlace Red Balls in Gaps: Now, place the 5 red balls in the 5 gaps created:B R _ _ _ R _ _ _ R _ R BCount the Arrangements: The blue balls can be arranged in 4! (4 factorial) ways. Each arrangement of blue balls allows for a unique placement of red balls.

Answered by benjaminforever12345 | 2024-10-20