A speech class has five freshmen and four sophomores. (Everyone is distinguishable.) In how many ways can they stand in line, so that at least four of the freshmen are standing next to each other?
+195 We can approach this problem by considering the possible arrangements of the freshmen and sophomores separately.
First, let's consider the arrangements of the freshmen. We want to count the number of ways that we can arrange the five freshmen such that at least four of them are standing next to each other. There are two cases to consider:
Case 1: Four freshmen are standing next to each other. There are 5 ways to choose which group of 4 freshmen will stand together, and 4! ways to arrange the 4 freshmen within that group. The remaining freshman can be placed in any of the remaining 6 positions. Therefore, there are a total of 5 × 4! × 6 = 720 ways to arrange the freshmen in this case.
Case 2: All five freshmen are standing together. There are 5! ways to arrange the five freshmen within the group. Therefore, there are a total of 5! = 120 ways to arrange the freshmen in this case.
Now, let's consider the arrangements of the sophomores. There are 4! ways to arrange the four sophomores.
To count the total number of arrangements that satisfy the given condition, we need to multiply the number of arrangements of the freshmen by the number of arrangements of the sophomores. Therefore, the total number of arrangements is:
(5 × 4! × 6) × 4! = 43,200
So there are 43,200 ways for the freshmen and sophomores to stand in line such that at least four of the freshmen are standing next to each other.
