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?

tomtom Feb 13, 2024

#1**+2 **

We treat the 4 freshmens __as a group. __

In the group of freshman there are 4! ways to order the freshmans.

To order the remaining objects, 4 sophomores, one freshman, and one group of freshmans, there is 6! ways to order this.

So there are a total of 4! * 6! ways to order

__But we must be careful, because if there is a group of 5 freshmanm, we overcount.__ The conbination S_{2}F_{1}F_{4}F_{3}F_{5}F_{2}S_{3}S_{4}S_{1}, can be formed as S_{2}F_{1} + a group + S_{3}S_{4}S_{1}, or S_{1} + a group + F_{2}S_{3}S_{4}S_{1}.

So we overcount each possiblity of five people in a group twice, so we need to subtract the number of possibilities with 5 people in a row.

The possibilities for 5 people in a row, treating the 5 freshmen as a group is 5! (the arrangments within the group) * (5!) (the ways to order 1 group + 4 sophomores).

So the answer should be 4!*6!-5!*5! = __2880 ways__

hairyberry Feb 13, 2024