+28 A speech class has 4 freshmen and 3 sophomores. (Everyone is distinguishable.) In how many ways can they stand in line, so that at least two of the freshmen are standing next to each other?
+399 Lets use complementary counting, it is much each to count all the cases where the freshman don't stand next to each other.
There are 7! ways to arrange 7 distinct people.
To have no two freshman stand next to each other, we must start with a freshman and alternate between freshman and sophomore like this:
F S F S F S F.
We can arrange the freshman in this arrangement 4! ways, and the sophomores 3! ways, so there are \(4!*3!\) ways for this arrangement.
Therefore there are 7!-3!*4! = 4896 ways to do this.
