Here's the question you clicked on:
Shadowys
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?
\[\frac{ (potato-1)! }{ potato!(potato-2)! }\]
Isn't that larger than 1?
Oh sorry that was number of ways it can be done.
The students can only be shuffled within their respective classrooms though...
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.
What was wrong with the answer I gave?
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.
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.
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.
@Shadowys can u PLZ explain the question once again.
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?
Then, what is wrong with 3!*3^12/3^15
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~
@satellite73 PLZ check this.
i don't understand the question
Oh my. Is it that vague?