A Discrete Plea!
The original situation: Six groups, each consisting of nine students, are in a theater. They sit in rows 1-6 and columns A-I, with all the students in the same group sitting in the same row.
Then, all the students leave the theater and return, again sitting in rows 1-6 and columns A-I, with all the students in the same group sitting in the same row.
What is the probability that:
- No group of students sits in its original row?
- Exactly two groups of students sit in their original row?
- At least three groups of students sit in their original row?
- No students sit in their original row or column?
Source: DiscreteMath Question posed on mathforum.org (9/22/2009) by "naslund19"
Hint: According to the poser of this problem, the key is to use the Principle of Inclusion-Exclusion.
If that does not help, you might try it with smaller cases, e.g. three groups of three students, etc....adjusting the number of rows and columns as needed.
Solution Commentary: When "naslund19" posed this question, only "Ben" responded (9/22/2009):
"Not sure what "inclusion-exclusion" has to do with this."
"With 6 rows, what is the probability that any one group will be in its original row? So what's the probability a group is not in its original row? So (1)-(3) and first part of (4) look like straight 'binomial' problems.
For the second part of (4), what is the probability that any one student is in his/her original column? So what's the probability (s)he is not in that column? How do we find the probability that EACH student is not in his/her original column?"
"Finally, how do we combine the two probabilities in (4) to get the probability no group is in its original row and no student is in her/his original column?"
"Thanks for a great problem!"
Now, that was Ben's response....do you agree?