A speech class has $4$ freshmen and $3$ sophomores. (Everyone is distinguishable.) In how many ways can they stand in line, so that all of the freshmen are standing next to each other?

magenta Jun 17, 2024

#1**+1 **

In order to solve this problem, we need to find how many ways we can arrange each age group.

The freshmen can stand in a line \(4! = 24\) ways.

As a group, they can arrange themselves in 4 ways across the line.

The sophoores can stand in a line \(3! = 6 \) ways.

Thus, we simply do

\(24 * 4 * 6 = 576 \)

So 576 is our final answer.

Thanks! :)

NotThatSmart Jun 17, 2024