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

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Question: How many ways can 3 red, 2 blue, and 1 green ball be arranged in a line?

Options:

Correct Answer: 60

Solution:

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

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

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

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: Understand that if all balls were different, we could arrange them in 6! (factorial of 6) ways. 6! = 720.
Step 3: Since some balls are the same (3 red and 2 blue), we need to divide by the factorials of the counts of each color to avoid overcounting.
Step 4: Calculate the factorials for the counts of each color: 3! for red balls, 2! for blue balls, and 1! for the green ball.
Step 5: Calculate 3! = 6, 2! = 2, and 1! = 1.
Step 6: Now, use the formula: Total arrangements = 6! / (3! * 2! * 1!).
Step 7: Substitute the values: Total arrangements = 720 / (6 * 2 * 1).
Step 8: Calculate the denominator: 6 * 2 * 1 = 12.
Step 9: Now divide: 720 / 12 = 60.
Step 10: Therefore, the total number of ways to arrange the balls is 60.

Permutations of Multisets – The question tests the understanding of how to calculate the number of distinct arrangements of items when there are repetitions, using the formula for permutations of multisets.