+0  
 
0
8
3
avatar+842 

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?

 Jun 17, 2024
 #1
avatar+1910 
+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! :)

 Jun 17, 2024

3 Online Users

avatar
avatar