In how many ways can 7 boys and 2 girls sit on a bench such that the girls do not sit next to each other?

There are 7 boys and 2 girls, making a total of $7+2=9$ people. The total number of ways to arrange 9 distinct people is $9!$.

$$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880$$

To find the arrangements where the two girls sit together, treat the two girls as a single unit. Now we have 7 boys and 1 unit of girls, making a total of $7+1=8$ units to arrange.
The number of ways to arrange these 8 units is $8!$.

$$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320$$

Within the unit of the two girls, the girls can arrange themselves in $2!$ ways (Girl 1 then Girl 2, or Girl 2 then Girl 1).

$$2! = 2 \times 1 = 2$$

So, the total number of ways the two girls sit together is the product of these two numbers:

$$8! \times 2! = 40,320 \times 2 = 80,640$$

The number of ways the girls do not sit next to each other is the total number of arrangements minus the number of arrangements where they do sit next to each other.

$$362,880 - 80,640 = 282,240$$

The number of ways they can arrange themselves so that the girls do not sit next to each other is $\boxed{\text{282,240}}$.