How many different ways can 3 red, 2 blue, and 1 green balls be arranged in a line?

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Q: How many different ways can 3 red, 2 blue, and 1 green balls be arranged in a line?

Solution: The total arrangements = 6! / (3! * 2! * 1!) = 60.

Steps: 6

Step 1: Count the total number of balls. We have 3 red, 2 blue, and 1 green ball. So, total balls = 3 + 2 + 1 = 6.
Step 2: Calculate the total arrangements if all balls were different. This is done by calculating 6! (6 factorial), which means 6 x 5 x 4 x 3 x 2 x 1 = 720.
Step 3: Since some balls are the same color, we need to divide by the arrangements of the identical balls. We have 3 red balls, which can be arranged among themselves in 3! (3 factorial) ways, which is 3 x 2 x 1 = 6.
Step 4: We also have 2 blue balls, which can be arranged among themselves in 2! (2 factorial) ways, which is 2 x 1 = 2.
Step 5: The green ball is unique, so it can be arranged in 1! (1 factorial) way, which is 1.
Step 6: Now, we calculate the total arrangements by dividing the total arrangements by the arrangements of identical balls: Total arrangements = 6! / (3! * 2! * 1!) = 720 / (6 * 2 * 1) = 720 / 12 = 60.