Here's another fun one! Fifteen students are distributed to three classes. Anong them, three of them are special students. What are the probability that each class has only one special student?

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\[\frac{ (potato-1)! }{ potato!(potato-2)! }\]

- anonymous

Potatoes!! lol

- anonymous

3! * 3^12

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- anonymous

Isn't that larger than 1?

- anonymous

Oh sorry that was number of ways it can be done.

- anonymous

(3! * 3^12)/3^15

- anonymous

The students can only be shuffled within their respective classrooms though...

- anonymous

And that means?

- anonymous

that means that once a special occupies a classroom, the other students might lose that chance to go into that classroom for the the probability that one student goes into one classroom.

- anonymous

What was wrong with the answer I gave?

- anonymous

your students were given three choices no matter the permutations of the three specials. however, in the case of two specials occupying two classrooms, they would only have one choice instead.

- anonymous

First special student have 3 choices
Second special student have 2 choices
Third special student have one choices.
And the rest students will have 3 choices.

- anonymous

yes.
But in the case of 1-1-13 classroom arrangement, where the first two classes will have the two specials, the 12 students don't have this choice of entering the classroom.

- anonymous

@Shadowys can u PLZ explain the question once again.

- anonymous

sorry~ ehheh
Fifteen students, three among them are special. are distributed to three classes, and each class must contain only one special student, there is no limit on the number of students in the class except that all students must go into a class. What is the probability?

- anonymous

Then, what is wrong with 3!*3^12/3^15

- anonymous

the students can be reshuffled in the classes without changing the class.
Frankly saying, I got your answer when I first did it before I tried simplifying the question and testing it again. Hmm~I'll check with teacher again. It might be that he's wrong~

- anonymous

@satellite73 PLZ check this.

- anonymous

i don't understand the question

- anonymous

Oh my. Is it that vague?

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