+1911 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?
+399 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 S2F1F4F3F5F2S3S4S1, can be formed as S2F1 + a group + S3S4S1, or S1 + a group + F2S3S4S1.
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
