In how many ways can 3 red, 2 blue, and 1 green balls be arranged in a line?
Practice Questions
1 question
Q1
In how many ways can 3 red, 2 blue, and 1 green balls be arranged in a line?
60
120
30
10
The total arrangements = 6! / (3! * 2! * 1!) = 60.
Questions & Step-by-step Solutions
1 item
Q
Q: In how many 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: 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!.
Step 3: Calculate 6!. This equals 720.
Step 4: Since some balls are identical, we need to divide by the factorial of the number of identical balls. We have 3 red balls, 2 blue balls, and 1 green ball.
Step 5: Calculate the factorials for the identical balls: 3! for red balls, 2! for blue balls, and 1! for green balls.
Step 6: Calculate 3! = 6, 2! = 2, and 1! = 1.
Step 7: Multiply the factorials of the identical balls: 3! * 2! * 1! = 6 * 2 * 1 = 12.
Step 8: Divide the total arrangements by the product of the factorials of the identical balls: 720 / 12 = 60.
Step 9: The final answer is 60, which means there are 60 different ways to arrange the balls.
