Home > Problem of the Week > Archive List > Detail

<< Prev 3/14/2010 Next >>

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:

  1. No group of students sits in its original row?
  2. Exactly two groups of students sit in their original row?
  3. At least three groups of students sit in their original row?
  4. 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?