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.