How many ways can 3 red balls and 2 blue balls be arranged in a line? (2016)

1 question

1 item

Q: How many ways can 3 red balls and 2 blue balls be arranged in a line? (2016)

Solution: The number of arrangements is 5! / (3! * 2!) = 10.

Steps: 8

Step 1: Count the total number of balls. We have 3 red balls and 2 blue balls, which gives us a total of 5 balls.
Step 2: Understand that we need to arrange these 5 balls in a line.
Step 3: If all the balls were different, we could arrange them in 5! (5 factorial) ways. 5! = 5 × 4 × 3 × 2 × 1 = 120.
Step 4: Since the red balls are identical (3 red balls) and the blue balls are identical (2 blue balls), we need to divide by the arrangements of the identical balls.
Step 5: Calculate the arrangements of the red balls, which is 3! (3 factorial). 3! = 3 × 2 × 1 = 6.
Step 6: Calculate the arrangements of the blue balls, which is 2! (2 factorial). 2! = 2 × 1 = 2.
Step 7: Now, we divide the total arrangements by the arrangements of the identical balls: 5! / (3! * 2!) = 120 / (6 * 2) = 120 / 12 = 10.
Step 8: Therefore, the number of ways to arrange 3 red balls and 2 blue balls in a line is 10.