In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?

To solve this problem, we can use the concept of permutations.

Step 1: First, let's consider the arrangement of the girls. Since there are 5 girls, we can arrange them in 5! (5 factorial) ways, which is equal to 5 x 4 x 3 x 2 x 1 = 120.

Step 2: Now, let's consider the arrangement of the boys. Since there are 3 boys, we can arrange them in 3! (3 factorial) ways, which is equal to 3 x 2 x 1 = 6.

Step 3: However, we need to ensure that no two boys are seated together. To achieve this, we can consider the girls and boys as separate entities and alternate their positions.

Step 4: We have 5 possible positions for the girls (denoted by G) and 4 possible positions for the boys (denoted by B). We can arrange them in the following pattern: GBGBGBG.

Step 5: Now, let's consider the arrangement of the girls within the G positions. Since there are 5 girls, we can arrange them in 5! ways within the G positions.

Step 6: Similarly, let's consider the arrangement of the boys within the B positions. Since there are 3 boys, we can arrange them in 3! ways within the B positions.

Step 7: Multiplying the number of arrangements of the girls (Step 5) and the number of arrangements of the boys (Step 6), we get 5! x 3! = 120 x 6 = 720.

Therefore, there are 720 ways to seat 5 girls and 3 boys in a row such that no two boys are together.

To solve this problem, we can use the concept of permutations.

Step 1: First, let's consider the arrangement of the girls. Since there are 5 girls, they can be seated in 5! (5 factorial) ways, which is equal to 120.

Step 2: Now, let's consider the arrangement of the boys. Since there are 3 boys, they can be seated in 3! (3 factorial) ways, which is equal to 6.

Step 3: However, we need to ensure that no two boys are seated together. To do this, we can consider the boys as "dividers" between the girls. We have 6 possible positions for the boys: before the first girl, between the first and second girl, between the second and third girl, between the third and fourth girl, between the fourth and fifth girl, and after the fifth girl.

Step 4: Now, we need to choose 3 out of the 6 positions for the boys. This can be done in 6 choose 3 ways, which is equal to 20.

Step 5: Finally, we multiply the number of ways to arrange the girls (120) by the number of ways to choose the positions for the boys (20) to get the total number of arrangements.

120 * 20 = 2400

Therefore, there are 2400 ways to seat 5 girls and 3 boys in a row such that no two boys are together.