JOhn had prepared the problem sets for Levels 1 to 5 of Geometry and Combinatorics for next month but forgot to label which set was for which level. Since Calvin didn't label them, the computer assigned them labels 1 through 5 randomly, with each label appearing only once. The probability that the problem sets given to each level are within one level of what they were supposed to be can be expressed as ab, where a and b are positive, coprime numbers. What is the value of a+b?

Question

JOhn had prepared the problem sets for Levels 1 to 5 of Geometry and Combinatorics for next month but forgot to label which set was for which level. Since Calvin didn't label them, the computer assigned them labels 1 through 5 randomly, with each label appearing only once. The probability that the problem sets given to each level are within one level of what they were supposed to be can be expressed as ab, where a and b are positive, coprime numbers. What is the value of a+b?

Answer 1

Details and assumptions

The computer randomly assigns each problem set to a level, and each level has exactly 1 problem set that is assigned. For example, the computer could assign the Level 1 problem set to Level 5 students, the Level 2 problem set to Level 4 students, the Level 3 problem set to Level 3 students, the Level 4 problem set to Level 2 students and the Level 5 problem set to Level 1 students.

Answer 2

To find the probability that the problem sets given to each level are within one level of what they were supposed to be, we need to consider the number of possible favorable outcomes (where the sets are within one level) and the total number of possible outcomes.

Let's examine the favorable outcomes first. Since none of the sets were labeled correctly, we can consider the possible scenarios where each set is consecutively off by one level:

1. If the set labeled as 1 is actually for Level 2, then there are two possibilities: the set labeled as 2 is either for Level 1 or for Level 3.
2. If the set labeled as 2 is actually for Level 1, then there are two possibilities: the set labeled as 1 is either for Level 2 or for Level 3.
3. If the set labeled as 3 is actually for Level 2, then there are three possibilities: the set labeled as 2 is either for Level 1, 3, or 4.
4. If the set labeled as 4 is actually for Level 3, then there are three possibilities: the set labeled as 3 is either for Level 2, 4, or 5.
5. If the set labeled as 5 is actually for Level 4, then there are two possibilities: the set labeled as 4 is either for Level 3 or for Level 5.

Combining these possibilities, we have a total of (2+2+3+3+2) = 12 favorable outcomes.

Now, let's consider the total number of possible outcomes. Since there are 5 sets to be labeled randomly, there are 5! = 5 x 4 x 3 x 2 x 1 = 120 possible ways to assign the labels to the sets.

Therefore, the probability of the problem sets given to each level being within one level is 12/120 = 1/10.

To express the probability as a fraction in the form ab, where a and b are coprime integers, we can simplify it further. In this case, a = 1 and b = 10.

Finally, the value of a + b = 1 + 10 = 11.

So, the value of a + b is 11.