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?

 Mar 21, 2024

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.

 Mar 21, 2024
edited by hairyberry  Mar 21, 2024

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