The exclamation point denotes the factorial (here is the Wikipedia page).
The answer to the question can be found as follows. We have a row of 7 chairs, $\Box\quad\Box\quad\Box\quad\Box\quad\Box\quad\Box\quad\Box$ The boys must form a block of 3 chairs. There are 5 such blocks: $\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box\quad\Box\quad\Box\quad\Box$ $\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box\quad\Box\quad\Box$ $\Box\quad\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box\quad\Box$ $\Box\quad\Box\quad\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare\quad\Box$ $\Box\quad\Box\quad\Box\quad\Box\quad\blacksquare\quad\blacksquare\quad\blacksquare$ So, when we tell the boys to sit, they must choose one of the 5 blocks of three chairs, and then choose how to arrange themselves in it. The first boy can choose any of the 3 chairs he wants, the second boy can choose from the 2 remaining chairs, and the last boy is forced to take whatever chair in the block is left. Thus, the number of ways of seating the boys is $5\times (3!)=5\times 3\times 2\times 1=30.$ After the boys are seated, the girls choose how to arrange themselves among the 4 remaining chairs. The first girl can choose any of the 4 chairs (etc.), so there are $4!=4\times 3\times 2\times 1=24$ ways the girls can be seated once the boys have chosen. Thus, overall there are $30\times 24=720=120\times 6=(5!)\times (3!)$ ways of seating these people.
As others have said, one can ask that the girls' chairs, and the block of chairs for the boys, be chosen together (in this way, we treat the group of boys as "another girl", for a total of 5 "girls" choosing among 5 "seats"); thus, we can have the boys choose one of the 5 "seats" (in reality, it is their block of chairs), the first girl chooses one of the 4 remaining chairs, the second girl chooses one of the 3 remaining chairs, etc. making for $5!=5\times 4\times 3\times 2\times 1=120$ ways of seating the girls and choosing the block of chairs for the boys. Then, all that remains is for the boys to arrange themselves within the block, which there are $3!=3\times 2\times 1=6$ ways to do, making (again) a total of $(5!)\times (3!)=120\times 6=720$ ways of arranging them.