Samir had prepared the problem tests for Stages 1 to 5 of Geometry and Combinatorics for next week but forgot to label which test was for which stage. Since Samir didn't label them, the computer assigned them labels 1 through 5 randomly, with each label appearing only once. The probability that the problems given to each stage are within one stage of what they were supposed to be can be expressed as a/b, where a and b are positive, coprime numbers. What is the value of a+b?

Question

Samir had prepared the problem tests for Stages 1 to 5 of Geometry and Combinatorics for next week but forgot to label which test was for which stage. Since Samir didn't label them, the computer assigned them labels 1 through 5 randomly, with each label appearing only once. The probability that the problems given to each stage are within one stage of what they were supposed to be can be expressed as a/b, where a and b are positive, coprime numbers. What is the value of a+b?

Details and assumptions
The computer randomly assigns each problem test to a stage, and each stage has exactly 1 problem test that is assigned. For example, the computer could assign the stage 1 problem test to stage 5 students, the stage 2 problem test to stage 4 students, the stage 3 problem test to stage 3 students, the stage 4 problem test to stage 2 students and the stage 5 problem test to stage 1 students.

Answer 1

To solve this problem, we need to count the number of favorable outcomes and the total number of possible outcomes, and then calculate the probability.

Let's consider each test separately and count the favorable outcomes for each test:

For the first test, there are two favorable outcomes: either it is assigned to the correct stage (stage 1), or it is assigned to the neighboring stage (stage 2).

For the second test, there are three favorable outcomes: it can be assigned to stage 1, stage 2, or stage 3.

Similarly, for the third, fourth, and fifth tests, there are three favorable outcomes each.

Now, let's count the total number of possible outcomes:

Since there are 5 tests, and each test can be assigned to one of the 5 stages, there are a total of 5^5 = 3125 possible outcomes.

To calculate the probability, we divide the number of favorable outcomes by the total number of possible outcomes:

Probability = (2 * 3 * 3 * 3 * 3) / 3125 = 54 / 3125

The fraction 54/3125 cannot be further simplified as both numbers are not divisible by any common factor. Therefore, the probability can be expressed as 54/3125.

Finally, we need to find the sum of the numerator and denominator: 54 + 3125 = 3179.

So, the value of a + b is 3179.