There are 5 boys and 4 girls in my class. All of them are distinguishable.
In how many ways can they be seated in a row of 9 chairs such that at least 2 boys are next to each other?
To find the number of ways they can be seated in a row of 9 chairs such that at least 2 boys are next to each other, we can use the principle of combinatorics.
First, let's find the total number of ways to arrange all 9 children in a row. We have 9 chairs and 9 children, so there are 9! (9 factorial) ways to arrange them.
Now, let's calculate the number of ways where no boys are next to each other. In this case, we need to alternate boys and girls in the seating arrangement. We have 5 boys and 4 girls, so the number of ways to arrange them in a row with no boys next to each other is the same as finding the number of ways to arrange the 4 girls in the 5 spaces between the boys. This can be calculated using the concept of permutation. We have 5 spaces and 4 girls, so the number of ways to arrange them is 5P4 = 5! / (5-4)! = 5! / 1! = 5!.
Finally, to find the number of ways where at least 2 boys are next to each other, we subtract the number of ways with no boys next to each other from the total number of arrangements. The number of ways with at least 2 boys next to each other is 9! - 5!.
Therefore, there are 9! - 5! = 362,880 - 120 = 362,760 ways to seat the children in a row of 9 chairs such that at least 2 boys are next to each other.