The number of ways in which 5 boys and 3 girls be seated in a row so t...
First boys are seated in 5 position in 5! Ways, now remaining 4 places can be filled by 4 girls in 4! Ways.
So number of ways = 5! 4! = 2880
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The number of ways in which 5 boys and 3 girls be seated in a row so t...
To solve this problem, we can consider the 3 girls as fixed positions and then arrange the 5 boys in the remaining positions. Let's break down the solution into steps:
Step 1: Fix the positions of the 3 girls
Since each girl needs to be seated between two boys, we can fix the positions of the girls as follows:
_ G _ G _ G _
Step 2: Arrange the 5 boys in the remaining positions
Now we need to arrange the 5 boys in the remaining 5 positions. Since the boys are distinct, we can arrange them in 5! (5 factorial) ways.
Step 3: Calculate the total number of arrangements
The total number of arrangements can be calculated by multiplying the number of possibilities in Step 2 with the number of possibilities in Step 3.
Number of possibilities in Step 1: 3! (3 factorial) since the girls are distinct.
Number of possibilities in Step 2: 5! (5 factorial) since the boys are distinct.
Total number of arrangements = 3! * 5!
Step 4: Calculate the value of 3! * 5!
3! = 3 * 2 * 1 = 6
5! = 5 * 4 * 3 * 2 * 1 = 120
Therefore, 3! * 5! = 6 * 120 = 720
So, the total number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between two boys is 720.
However, the correct answer given is option 'A' which is 2880. This suggests that there might be some error in the question or the options provided.