In how many ways can 4 boys and 3 girl be seated in a row of 7 chairs if boys and girls alternate?
To seat 4 boys and 3 girls in a row of 7 chairs so that boys and girls alternate, we can consider the boys and girls as separate groups and then arrange them.
First, arrange the 4 boys in the 4 available chairs: 4 boys in 4 chairs can be arranged in 4! = 24 ways.
Next, arrange the 3 girls in the 3 available chairs: 3 girls in 3 chairs can be arranged in 3! = 6 ways.
Since the boys and girls must alternate, the total number of ways to arrange them is 24 * 6 = <<24*6=144>>144 ways.
To determine the number of ways the boys and girls can be seated in alternating positions, we can treat the boys and girls as distinct groups and consider their arrangements separately.
First, let's arrange the boys in the chairs. Since there are 4 boys, we can arrange them in 4! (4 factorial) ways.
Next, let's arrange the girls in the remaining chairs. Since there are 3 girls, we can arrange them in 3! (3 factorial) ways.
Now, to get the total number of ways for alternate seating, we multiply the number of ways to arrange the boys by the number of ways to arrange the girls.
Total number of ways = (4!) * (3!) = 24 * 6 = 144
Therefore, there are 144 ways to seat 4 boys and 3 girls in a row of 7 chairs if they alternate.
To solve this problem, we can treat the arrangement of boys and girls as a single sequence.
Step 1: Place the boys and girls in alternate positions. We have 4 boys and 3 girls, so we can arrange them in (4+3)! = 7! ways.
Step 2: Within the group of boys and girls, we can arrange the boys among themselves and the girls among themselves. The boys can be arranged in 4! ways and the girls can be arranged in 3! ways.
Step 3: Multiply the results from steps 1 and 2 to get the total number of arrangements.
Total number of arrangements = (4+3)! * 4! * 3! = 7! * 4! * 3!
Therefore, there are 7! * 4! * 3! = 5040 * 24 * 6 = 725,760 ways to seat the 4 boys and 3 girls in a row of 7 chairs if boys and girls alternate.