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

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

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

Steps: 9

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 using the factorial of the total number of balls, which is 6! (6 factorial).
Step 3: Calculate 6!. This means 6 x 5 x 4 x 3 x 2 x 1 = 720.
Step 4: Since some balls are the same color, we need to divide by the factorial of the number of each color of balls to avoid counting duplicates. We have 3 red balls, 2 blue balls, and 1 green ball.
Step 5: Calculate the factorial for each color: 3! for red balls, 2! for blue balls, and 1! for green balls.
Step 6: Calculate 3! = 3 x 2 x 1 = 6, 2! = 2 x 1 = 2, and 1! = 1.
Step 7: Multiply these factorials together: 3! * 2! * 1! = 6 * 2 * 1 = 12.
Step 8: Now, divide the total arrangements by the product of the factorials: 720 / 12 = 60.
Step 9: The final answer is that there are 60 different ways to arrange the 3 red, 2 blue, and 1 green balls in a row.