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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
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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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