Permuted Problem Sets

[Read Joshua's second solution in the comments. - Calvin]

There are several cases to consider when working with this type of problem. First, we notice that in order for the 1 to be placed within one level of where it is supposed to be, the computer must assign it to the level 1 or 2 problem. Similarly, the 5 must be assigned to the level 4 or 5 problem. These are the most restricted numbers, so we have motivation to assign cases to the different configurations of 1's and 5's. We may proceed as follows:

Case 1: 1 is assigned to 2, 5 is assigned to 4 We see that in this configuration, we force the computer to assign the 4 to the level 5 and the 2 to level 1 to meet our specifications. This leaves only one spot for the 3 (on level 3), so this totals to $1$ configuration.

Case 2: 1 is assigned to 2, 5 is assigned to 5 In this case, we force the computer to assign the 2 to level 1. This leaves two spots for the 3 and the 4, and these may be assigned in $2!=2$ ways, so this totals to $2$ configurations.

Case 3: 1 is assigned to 1, 5 is assigned to 4 By symmetry, this case is identical to the one above (just replace each assignment number $n$ and level $l$ by $6-n$ and $6-l$ , respectively), so there are also $2$ configurations.

Case 4: 1 is assigned to 1, 5 is assigned to 5 In this case we are left with the 2, 3, and 4 left to be assigned to levels 2, 3, and 4. We note that once the computer assigns a 3 to any of these numbers, the rest of the configuration is uniquely determined. Since the 3 can be in three places, there are a total of $3$ configurations.

These cases cover all of the configurations with no repeats. There are $5!$ ways for the computer to assign 5 numbers to 5 levels, so the probability that the computer assigns the numbers to levels within $1$ of the actual value is $\dfrac{1+2+2+3}{5!}=\dfrac{8}{120}=\dfrac{1}{15}$ , so the answer is $1+15=\boxed{16}$

Total possibility will be 5 combination 1, which is 5!. (Because it done by the 5 spaces, 4 spaces...). but in the way to get the within one lever which means ±1. so just do it because if we limit in the number 1, only two trials are required.

1-2-3-4-5 1-2-3-5-4 1-2-4-3-5 1-3-2-4-5 1-3-2-5-4 2-1-3-4-5 2-1-3-5-4 2-1-4-3-5 (in total 8 can be done) so, 8/120= 1/15. a=1, b=15 a+b=16

Instead of calculating a probability directly, we will count the number of total outcomes and the number of favourable outcomes. There are $5! = 120$ different ways that the problems could be assigned. Let $s_n$ be the number of ways we can assign $n$ levels of problems to $n$ levels of students so that the level of problems assigned to each level of students is within one. If we consider the highest level of students, then they can either receive the highest level of problems, or they can receive the second highest level of problems.

Case 1. If they receive the highest level of problems, then the number of ways to assign the rest of the levels is $s_{n-1}$ .

Case 2. If they receive the second highest level of problems, then the highest level of problems must be assigned to the second highest level of students, so there are $s_{n-2}$ ways to assign the remaining problem levels.

Thus, $s_n$ satisfies the recurrence relationship $s_n = s_{n-1} + s_{n-2}$ . If there is only one level, then there is only 1 way to assign the levels, and if there are two levels there will be 2 ways of assigning the levels. Thus $s_1 = 1$ and $s_2 = 2$ and we see that the recurrence is just the Fibonacci sequence, so there will be $s_5 = 8$ ways to assign the problem levels to meet the desired criteria. Thus, the probability is $\frac{8}{120} = \frac{1}{15}$ . Hence $a + b = 1 + 15 = 16$ .