In how many ways can 7 boys and 2 girls be arranged in a row such that no two girls sit together?

There are 7 boys. The number of ways to arrange these 7 boys on a bench is $7!$.

When the 7 boys are arranged, they create 8 possible spaces (gaps) where the girls can sit so that no two girls are next to each other.
Let 'B' represent a boy and '_' represent a possible position for a girl:

There are 8 such positions.

We have 2 girls to place in 8 distinct positions. Since the girls are distinct, the order in which they are placed matters. This is a permutation problem.
The number of ways to choose 2 positions out of 8 and arrange the 2 girls in them is given by $P(8, 2)$.

To find the total number of ways, multiply the number of ways to arrange the boys by the number of ways to place the girls.
Total ways = (Ways to arrange boys) $\times$ (Ways to place girls)
Total ways = $7! \times P(8, 2)$
Total ways = $5,040 \times 56$
Total ways = 282,240