A small theater only have $10$ numbered seats. However there is a row of $30$ people to enter into the theater. A group of $3$ friends want to enter into the theater in the same movie session. In how many different ways the $30$ can enter into the theater in order to the $3$ friends could enter in the same session together?
Well, if the theater only have $10$ seats, and there are $30$ people to see the movie. There should be $3$ sessions. As the seats are numbered, in each session the $3$ friends can be sit in $Pr(10,3)$ different ways. But there will be $3$ sessions, they can enter in the $1st$, $2nd$ or in the $3th$ session. So there are $3 \cdot Pr(10,3)$.
But the book's solution is $3! \cdot 10 \cdot 27!$, that I don't understand. Can you help me?Thanks